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#include <bits/stdc++.h> int main() { int a, b, c, t, i, n; scanf("%d %d %d", &a, &b, &c); if ((a + b) > c && (a + c) > b && (b + c) > a) t = 0; else if ((a + b) <= c) t = c - (a + b) + 1; else if ((a + c) <= b) t = b - (a + c) + 1; else t = a - (b + c) + 1; printf("%d\n", t); return 0; }
### Prompt Generate a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> int main() { int a, b, c, t, i, n; scanf("%d %d %d", &a, &b, &c); if ((a + b) > c && (a + c) > b && (b + c) > a) t = 0; else if ((a + b) <= c) t = c - (a + b) + 1; else if ((a + c) <= b) t = b - (a + c) + 1; else t = a - (b + c) + 1; printf("%d\n", t); return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c; int arr[3]; for (int i = 0; i < 3; i++) { cin >> arr[i]; } int len = sizeof(arr) / sizeof(arr[0]); sort(arr, arr + len); if ((arr[len - 3] + arr[len - 2]) > arr[len - 1]) cout << "0" << endl; else cout << (arr[len - 1] - (arr[len - 3] + arr[len - 2])) + 1 << endl; }
### Prompt Please provide a CPP coded solution to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c; int arr[3]; for (int i = 0; i < 3; i++) { cin >> arr[i]; } int len = sizeof(arr) / sizeof(arr[0]); sort(arr, arr + len); if ((arr[len - 3] + arr[len - 2]) > arr[len - 1]) cout << "0" << endl; else cout << (arr[len - 1] - (arr[len - 3] + arr[len - 2])) + 1 << endl; } ```
#include <bits/stdc++.h> using namespace std; int t[3]; int main() { cin >> t[0] >> t[1] >> t[2]; sort(t, t + 3); if (t[0] + t[1] >= t[2] + 1) cout << 0; else cout << t[2] - t[1] - t[0] + 1; return 0; }
### Prompt Generate a cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int t[3]; int main() { cin >> t[0] >> t[1] >> t[2]; sort(t, t + 3); if (t[0] + t[1] >= t[2] + 1) cout << 0; else cout << t[2] - t[1] - t[0] + 1; return 0; } ```
#include <bits/stdc++.h> using namespace std; int a[3], sol; int main() { cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); int x = a[0] + a[1]; while (x <= a[2]) { x++; sol++; } cout << sol; return 0; }
### Prompt In CPP, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int a[3], sol; int main() { cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); int x = a[0] + a[1]; while (x <= a[2]) { x++; sol++; } cout << sol; return 0; } ```
#include <bits/stdc++.h> using namespace std; bool Trangle(double a, double b, double c) { return a + b > c && a + c > b && b + c > a && a - b < c && a - c < b && b - c < a; } int main() { int num[3]; int ans(0); cin >> num[0] >> num[1] >> num[2]; { ans = 0; sort(num, num + 3); while (!Trangle(num[0], num[1], num[2])) { if (ans % 2) num[1]++; else num[0]++; ans++; } cout << ans << endl; } }
### Prompt Please create a solution in cpp to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; bool Trangle(double a, double b, double c) { return a + b > c && a + c > b && b + c > a && a - b < c && a - c < b && b - c < a; } int main() { int num[3]; int ans(0); cin >> num[0] >> num[1] >> num[2]; { ans = 0; sort(num, num + 3); while (!Trangle(num[0], num[1], num[2])) { if (ans % 2) num[1]++; else num[0]++; ans++; } cout << ans << endl; } } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[3]; int n, i; for (i = 0; i < 3; i++) cin >> a[i]; sort(a, a + 3); if (a[0] + a[1] > a[2]) cout << "0" << "\n"; else cout << (a[2] - a[1] - a[0]) + 1 << "\n"; return 0; }
### Prompt Develop a solution in cpp to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[3]; int n, i; for (i = 0; i < 3; i++) cin >> a[i]; sort(a, a + 3); if (a[0] + a[1] > a[2]) cout << "0" << "\n"; else cout << (a[2] - a[1] - a[0]) + 1 << "\n"; return 0; } ```
#include <bits/stdc++.h> int main() { int a, b, c, d, e, f, g, h = 0, i = 0; scanf("%d%d%d", &a, &b, &c); if (a >= b) { if (a + b <= c) { h = h + c - a - b + 1; b = b + h; } } else if (a < b) { if (a + b <= c) { h = h + c - a - b + 1; a = a + h; } } if (b >= c) { if (c + b <= a) { h = h + a - c - b + 1; c = c + h; } } else if (b < c) { if (c + b <= a) { h = h + a - c - b + 1; b = b + h; } } if (a + c <= b) { h = h + b - a - c + 1; } printf("%d", h); }
### Prompt Develop a solution in cpp to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> int main() { int a, b, c, d, e, f, g, h = 0, i = 0; scanf("%d%d%d", &a, &b, &c); if (a >= b) { if (a + b <= c) { h = h + c - a - b + 1; b = b + h; } } else if (a < b) { if (a + b <= c) { h = h + c - a - b + 1; a = a + h; } } if (b >= c) { if (c + b <= a) { h = h + a - c - b + 1; c = c + h; } } else if (b < c) { if (c + b <= a) { h = h + a - c - b + 1; b = b + h; } } if (a + c <= b) { h = h + b - a - c + 1; } printf("%d", h); } ```
#include <bits/stdc++.h> using namespace std; int sides[3]; int main() { for (int i = 0; i < 3; ++i) scanf("%d", &sides[i]); sort(sides, sides + 3); if (sides[0] + sides[1] > sides[2]) printf("0"); else printf("%d", sides[2] + 1 - (sides[0] + sides[1])); return 0; }
### Prompt Develop a solution in CPP to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int sides[3]; int main() { for (int i = 0; i < 3; ++i) scanf("%d", &sides[i]); sort(sides, sides + 3); if (sides[0] + sides[1] > sides[2]) printf("0"); else printf("%d", sides[2] + 1 - (sides[0] + sides[1])); return 0; } ```
#include <bits/stdc++.h> using namespace std; const long long MAXN = 2005; const long long INF = 0x3f3f3f3f; const long long LINF = 0x1f3f3f3f3f3f3f3f; const long long MOD = (long long)1e9 + 7; const long long OVER_FLOW = 0xffffffff; long long s, T, n, m, M, k, A, B, p; long long buf[MAXN]; int main() { ios::sync_with_stdio(false); cin >> buf[1] >> buf[2] >> buf[3]; sort(buf + 1, buf + 1 + 3); cout << max(0ll, buf[3] - buf[2] - buf[1] + 1) << endl; return 0; }
### Prompt Your challenge is to write a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; const long long MAXN = 2005; const long long INF = 0x3f3f3f3f; const long long LINF = 0x1f3f3f3f3f3f3f3f; const long long MOD = (long long)1e9 + 7; const long long OVER_FLOW = 0xffffffff; long long s, T, n, m, M, k, A, B, p; long long buf[MAXN]; int main() { ios::sync_with_stdio(false); cin >> buf[1] >> buf[2] >> buf[3]; sort(buf + 1, buf + 1 + 3); cout << max(0ll, buf[3] - buf[2] - buf[1] + 1) << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; map<char, int> m; map<char, int> m1; int main() { long long int n = 0, d, i = 0, j = 0, l = 0, r = 0, a = 0, b = 0, c1, c2, p, b2, s1 = 0, t = 0, t1 = 0, y = 0, x = 0, s2 = 0, c = 0; char ch; cin >> a >> b >> c; vector<int> adj[100001]; int s[3]; s[0] = a; s[1] = b; s[2] = c; n = 3; sort(s, s + n); a = s[2]; b = s[1]; c = s[0]; if (a >= (b + c)) { while (a >= b + c) { t++; c++; } } if (b >= a + c) { while (b >= a + c) { t++; c++; } } if (c >= b + a) { while (c >= b + a) { t++; b++; } } cout << t << endl; return 0; }
### Prompt In Cpp, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; map<char, int> m; map<char, int> m1; int main() { long long int n = 0, d, i = 0, j = 0, l = 0, r = 0, a = 0, b = 0, c1, c2, p, b2, s1 = 0, t = 0, t1 = 0, y = 0, x = 0, s2 = 0, c = 0; char ch; cin >> a >> b >> c; vector<int> adj[100001]; int s[3]; s[0] = a; s[1] = b; s[2] = c; n = 3; sort(s, s + n); a = s[2]; b = s[1]; c = s[0]; if (a >= (b + c)) { while (a >= b + c) { t++; c++; } } if (b >= a + c) { while (b >= a + c) { t++; c++; } } if (c >= b + a) { while (c >= b + a) { t++; b++; } } cout << t << endl; return 0; } ```
#include <bits/stdc++.h> int a[4]; int main() { scanf("%d%d%d", &a[1], &a[2], &a[3]); std::sort(a + 1, a + 3 + 1); if (a[1] + a[2] > a[3]) { printf("%d\n", 0); } else { printf("%d\n", a[3] - a[1] - a[2] + 1); } return 0; }
### Prompt Your task is to create a cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> int a[4]; int main() { scanf("%d%d%d", &a[1], &a[2], &a[3]); std::sort(a + 1, a + 3 + 1); if (a[1] + a[2] > a[3]) { printf("%d\n", 0); } else { printf("%d\n", a[3] - a[1] - a[2] + 1); } return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); if (a[0] + a[1] > a[2]) cout << 0; else cout << a[2] - a[0] - a[1] + 1; }
### Prompt Create a solution in cpp for the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); if (a[0] + a[1] > a[2]) cout << 0; else cout << a[2] - a[0] - a[1] + 1; } ```
#include <bits/stdc++.h> using namespace std; int a[10]; int main() { cin >> a[1] >> a[2] >> a[3]; sort(a + 1, a + 1 + 3); cout << max(a[3] - a[1] - a[2] + 1, 0) << endl; return 0; }
### Prompt Create a solution in Cpp for the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int a[10]; int main() { cin >> a[1] >> a[2] >> a[3]; sort(a + 1, a + 1 + 3); cout << max(a[3] - a[1] - a[2] + 1, 0) << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[4]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3, greater<int>()); if (a[0] < a[1] + a[2]) cout << "0" << endl; else cout << a[0] - a[1] - a[2] + 1 << endl; return 0; }
### Prompt Create a solution in cpp for the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[4]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3, greater<int>()); if (a[0] < a[1] + a[2]) cout << "0" << endl; else cout << a[0] - a[1] - a[2] + 1 << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main(int argc, const char* argv[]) { int leg[3]; int m, n; int x; for (int i = 0; i < 3; i++) { cin >> leg[i]; } m = 0; n = 0; for (int i = 0; i < 3; i++) { if (leg[i] >= m) { n = i; m = leg[i]; } } m = 0; for (int i = 0; i < 3; i++) { if (i == n) { continue; } m += leg[i]; } x = 0; if (leg[n] >= m) { x = leg[n] - m + 1; } cout << x; return 0; }
### Prompt Please provide a Cpp coded solution to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main(int argc, const char* argv[]) { int leg[3]; int m, n; int x; for (int i = 0; i < 3; i++) { cin >> leg[i]; } m = 0; n = 0; for (int i = 0; i < 3; i++) { if (leg[i] >= m) { n = i; m = leg[i]; } } m = 0; for (int i = 0; i < 3; i++) { if (i == n) { continue; } m += leg[i]; } x = 0; if (leg[n] >= m) { x = leg[n] - m + 1; } cout << x; return 0; } ```
#include <bits/stdc++.h> using namespace std; const long long int inf = 1e13 + 7; void solve(); void sieve(); int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout << fixed; cout << setprecision(10); long long int t = 1; while (t--) solve(); return 0; } struct point { long long int x, y; point operator+(const point& a) { point res; res.x = x + a.x; res.y = y + a.y; return res; } point operator-(const point& a) { point res; res.x = x - a.x; res.y = y - a.y; return res; } long long int operator*(const point& a) { long long int res; res = x * a.y - y * a.x; return res; } long long int operator|(const point& a) { long long int res; res = x * a.x + y * a.y; return res; } }; long long int orientation(point p1, point p2, point p3) { return (p1 - p2) * (p1 - p3); } void convexhull(vector<long long int>& v) { ; } long double dist(point p1, point p2) { return sqrtl((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y)); } void solve() { long long int a, b, c; cin >> a >> b >> c; cout << max(0LL, 1 + 2 * max(a, max(b, c)) - (a + b + c)); }
### Prompt Please provide a CPP coded solution to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; const long long int inf = 1e13 + 7; void solve(); void sieve(); int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout << fixed; cout << setprecision(10); long long int t = 1; while (t--) solve(); return 0; } struct point { long long int x, y; point operator+(const point& a) { point res; res.x = x + a.x; res.y = y + a.y; return res; } point operator-(const point& a) { point res; res.x = x - a.x; res.y = y - a.y; return res; } long long int operator*(const point& a) { long long int res; res = x * a.y - y * a.x; return res; } long long int operator|(const point& a) { long long int res; res = x * a.x + y * a.y; return res; } }; long long int orientation(point p1, point p2, point p3) { return (p1 - p2) * (p1 - p3); } void convexhull(vector<long long int>& v) { ; } long double dist(point p1, point p2) { return sqrtl((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y)); } void solve() { long long int a, b, c; cin >> a >> b >> c; cout << max(0LL, 1 + 2 * max(a, max(b, c)) - (a + b + c)); } ```
#include <bits/stdc++.h> int main() { int a, b, c, d, t = 0; scanf("%d %d %d", &a, &b, &c); if (a > b) { if (a > c) { d = b + c; if (d > a) t = 0; else { while (1) { t++; d++; if (d > a) break; } } } } if (b > c) { if (b > a) { d = a + c; if (d > b) t = 0; else { while (1) { t++; d++; if (d > b) break; } } } } if (c > a) { if (c > b) { d = a + b; if (d > c) t = 0; else { while (1) { t++; d++; if (d > c) break; } } } } printf("%d\n", t); return 0; }
### Prompt Develop a solution in cpp to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> int main() { int a, b, c, d, t = 0; scanf("%d %d %d", &a, &b, &c); if (a > b) { if (a > c) { d = b + c; if (d > a) t = 0; else { while (1) { t++; d++; if (d > a) break; } } } } if (b > c) { if (b > a) { d = a + c; if (d > b) t = 0; else { while (1) { t++; d++; if (d > b) break; } } } } if (c > a) { if (c > b) { d = a + b; if (d > c) t = 0; else { while (1) { t++; d++; if (d > c) break; } } } } printf("%d\n", t); return 0; } ```
#include <bits/stdc++.h> using namespace std; signed main() { vector<unsigned> a(3); for (unsigned& x : a) cin >> x; sort(a.begin(), a.end()); if (a[2] < a[0] + a[1]) { cout << 0; exit(0); } cout << min(a[2] - (a[1] + a[0]), min(a[1] - (a[2] - a[0]), a[0] - (a[2] - a[0]))) + 1; }
### Prompt Please formulate a cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; signed main() { vector<unsigned> a(3); for (unsigned& x : a) cin >> x; sort(a.begin(), a.end()); if (a[2] < a[0] + a[1]) { cout << 0; exit(0); } cout << min(a[2] - (a[1] + a[0]), min(a[1] - (a[2] - a[0]), a[0] - (a[2] - a[0]))) + 1; } ```
#include <bits/stdc++.h> using namespace std; int main() { vector<int> a(3, 0); cin >> a[0] >> a[1] >> a[2]; sort(a.begin(), a.end()); int cnt = 0; while (a[0] + a[1] <= a[2]) { a[0]++; cnt++; if (a[0] > a[1]) { swap(a[0], a[1]); } } cout << cnt << endl; return 0; }
### Prompt Generate a cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { vector<int> a(3, 0); cin >> a[0] >> a[1] >> a[2]; sort(a.begin(), a.end()); int cnt = 0; while (a[0] + a[1] <= a[2]) { a[0]++; cnt++; if (a[0] > a[1]) { swap(a[0], a[1]); } } cout << cnt << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int arr[3]; for (int i = 0; i < 3; i++) cin >> arr[i]; sort(arr, arr + 3); int ans = arr[2] - arr[1] - arr[0]; cout << (ans < 0 ? 0 : ans + 1) << endl; }
### Prompt Generate a cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int arr[3]; for (int i = 0; i < 3; i++) cin >> arr[i]; sort(arr, arr + 3); int ans = arr[2] - arr[1] - arr[0]; cout << (ans < 0 ? 0 : ans + 1) << endl; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c; cin >> a >> b >> c; int jum = a + b + c; int maks = max(max(a, b), c); int ans = maks - (jum - maks) + 1; if (ans < 1) cout << 0 << endl; else cout << ans << endl; }
### Prompt In Cpp, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c; cin >> a >> b >> c; int jum = a + b + c; int maks = max(max(a, b), c); int ans = maks - (jum - maks) + 1; if (ans < 1) cout << 0 << endl; else cout << ans << endl; } ```
#include <bits/stdc++.h> int in() { int x = 0; char c = getchar(); bool neg = false; while (c != '-' && ('0' > c || c > '9')) c = getchar(); if (c == '-') neg = true, c = getchar(); while ('0' <= c && c <= '9') x = 10 * x + c - '0', c = getchar(); if (neg) x = -x; return x; } void out(int x) { if (x > 9) out(x / 10); putchar(x % 10 + '0'); } using namespace std; int main() { int kq = 0; int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); while (a[0] + a[1] + kq <= a[2]) kq++; cout << kq; }
### Prompt Your challenge is to write a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> int in() { int x = 0; char c = getchar(); bool neg = false; while (c != '-' && ('0' > c || c > '9')) c = getchar(); if (c == '-') neg = true, c = getchar(); while ('0' <= c && c <= '9') x = 10 * x + c - '0', c = getchar(); if (neg) x = -x; return x; } void out(int x) { if (x > 9) out(x / 10); putchar(x % 10 + '0'); } using namespace std; int main() { int kq = 0; int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); while (a[0] + a[1] + kq <= a[2]) kq++; cout << kq; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[5]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); if (a[0] + a[1] > a[2]) { cout << 0 << endl; } else { int cnt = 0; while (a[0] + a[1] <= a[2]) { a[1]++; cnt++; } cout << cnt << endl; } return 0; }
### Prompt Develop a solution in CPP to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[5]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); if (a[0] + a[1] > a[2]) { cout << 0 << endl; } else { int cnt = 0; while (a[0] + a[1] <= a[2]) { a[1]++; cnt++; } cout << cnt << endl; } return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c; cin >> a >> b >> c; if (a >= b && a >= c) { if (b + c > a) cout << "0"; else cout << a - (b + c) + 1; } else if (b >= a && b >= c) { if (a + c > b) cout << "0"; else cout << b - (a + c) + 1; } else { if (a + b > c) cout << "0"; else cout << c - (a + b) + 1; } }
### Prompt Please provide a CPP coded solution to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c; cin >> a >> b >> c; if (a >= b && a >= c) { if (b + c > a) cout << "0"; else cout << a - (b + c) + 1; } else if (b >= a && b >= c) { if (a + c > b) cout << "0"; else cout << b - (a + c) + 1; } else { if (a + b > c) cout << "0"; else cout << c - (a + b) + 1; } } ```
#include <bits/stdc++.h> using namespace std; int arr[4], x; int main() { for (int i = 0; i < 3; i++) { cin >> x; arr[i] = x; } for (int i = 0; i < 3; i++) { for (int j = 0; j < 3 - i - 1; j++) if (arr[j] > arr[j + 1]) swap(arr[j], arr[j + 1]); } int s = (arr[2] + 1) - (arr[0] + arr[1]); if (s <= 0) cout << "0"; else cout << s; return 0; }
### Prompt Please provide a Cpp coded solution to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int arr[4], x; int main() { for (int i = 0; i < 3; i++) { cin >> x; arr[i] = x; } for (int i = 0; i < 3; i++) { for (int j = 0; j < 3 - i - 1; j++) if (arr[j] > arr[j + 1]) swap(arr[j], arr[j + 1]); } int s = (arr[2] + 1) - (arr[0] + arr[1]); if (s <= 0) cout << "0"; else cout << s; return 0; } ```
#include <bits/stdc++.h> using namespace std; long long INF = 9223372036854775807; int md = 100000000; const int WHITE = 0; const int BLACK = 1; int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); vector<int> m; int a; cin >> a; m.push_back(a); cin >> a; m.push_back(a); cin >> a; m.push_back(a); sort(m.begin(), m.end()); if (m[0] + m[1] <= m[2]) cout << m[2] - m[1] - m[0] + 1 << endl; else cout << 0 << endl; return 0; }
### Prompt In cpp, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; long long INF = 9223372036854775807; int md = 100000000; const int WHITE = 0; const int BLACK = 1; int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); vector<int> m; int a; cin >> a; m.push_back(a); cin >> a; m.push_back(a); cin >> a; m.push_back(a); sort(m.begin(), m.end()); if (m[0] + m[1] <= m[2]) cout << m[2] - m[1] - m[0] + 1 << endl; else cout << 0 << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { long long a, b, c, m, k; cin >> a >> b >> c; m = max(max(a, b), c); k = min(min(a, b), c); b = a + b + c - m - k; a = m; c = k; m = 0; cout << max(m, a - b - c + 1); }
### Prompt Create a solution in Cpp for the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { long long a, b, c, m, k; cin >> a >> b >> c; m = max(max(a, b), c); k = min(min(a, b), c); b = a + b + c - m - k; a = m; c = k; m = 0; cout << max(m, a - b - c + 1); } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); cout << max(0, a[2] - (a[1] + a[0]) + 1) << "\n"; }
### Prompt Your task is to create a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); cout << max(0, a[2] - (a[1] + a[0]) + 1) << "\n"; } ```
#include <bits/stdc++.h> using namespace std; int main() { int ar[3], sm = 0, i, d; for (i = 0; i < 3; i++) scanf("%d", &ar[i]); d = 0; sort(ar, ar + 3); sm = ar[0] + ar[1]; if (ar[2] >= sm) { d = ar[2] - sm; d++; } printf("%d\n", d); return 0; }
### Prompt Please formulate a cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int ar[3], sm = 0, i, d; for (i = 0; i < 3; i++) scanf("%d", &ar[i]); d = 0; sort(ar, ar + 3); sm = ar[0] + ar[1]; if (ar[2] >= sm) { d = ar[2] - sm; d++; } printf("%d\n", d); return 0; } ```
#include <bits/stdc++.h> int main() { long long int a, b, c, count, A, B, C; scanf("%lld %lld %lld", &a, &b, &c); count = 0; while ((a + b) <= c || (b + c) <= a || (a + c) <= b) { A = a; B = b; C = c; if (A <= B && A <= C) { a++; count++; } if ((a + b) <= c || (b + c) <= a || (a + c) <= b) { A = a; B = b; C = c; if (B <= A && B <= C) { b++; count++; } } if ((a + b) <= c || (b + c) <= a || (a + c) <= b) { A = a; B = b; C = c; if (C <= A && C <= B) { c++; count++; } } } printf("%lld\n", count); return 0; }
### Prompt In CPP, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> int main() { long long int a, b, c, count, A, B, C; scanf("%lld %lld %lld", &a, &b, &c); count = 0; while ((a + b) <= c || (b + c) <= a || (a + c) <= b) { A = a; B = b; C = c; if (A <= B && A <= C) { a++; count++; } if ((a + b) <= c || (b + c) <= a || (a + c) <= b) { A = a; B = b; C = c; if (B <= A && B <= C) { b++; count++; } } if ((a + b) <= c || (b + c) <= a || (a + c) <= b) { A = a; B = b; C = c; if (C <= A && C <= B) { c++; count++; } } } printf("%lld\n", count); return 0; } ```
#include <bits/stdc++.h> using namespace std; const long long MaxN = 5; long long n, a[MaxN], f[MaxN][13], cs; bool kt; void work() { sort(a + 1, a + 4); if (a[1] + a[2] > a[3]) cout << 0; else cout << a[3] - a[2] - a[1] + 1; } void nhap() { cin >> a[1] >> a[2] >> a[3]; } int main() { nhap(); work(); }
### Prompt Please provide a cpp coded solution to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; const long long MaxN = 5; long long n, a[MaxN], f[MaxN][13], cs; bool kt; void work() { sort(a + 1, a + 4); if (a[1] + a[2] > a[3]) cout << 0; else cout << a[3] - a[2] - a[1] + 1; } void nhap() { cin >> a[1] >> a[2] >> a[3]; } int main() { nhap(); work(); } ```
#include <bits/stdc++.h> using namespace std; template <class T1> void deb(T1 e1) { cout << e1 << endl; } template <class T1, class T2> void deb(T1 e1, T2 e2) { cout << e1 << " " << e2 << endl; } template <class T1, class T2, class T3> void deb(T1 e1, T2 e2, T3 e3) { cout << e1 << " " << e2 << " " << e3 << endl; } void check_time() {} const int res = 1e5 + 5; const long long int mod = 1e9 + 7; int main() { int a, b, c, maxi = 1e9; scanf("%d %d %d", &a, &b, &c); if (a + b > c and a + c > b and b + c > a) { deb("0"); return 0; } if (a + b <= c) { maxi = min(maxi, c - a - b + 1); } if (a + c <= b) { maxi = min(maxi, b - a - c + 1); } if (b + c <= a) { maxi = min(maxi, a - c - b + 1); } deb(maxi); return 0; }
### Prompt Develop a solution in CPP to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; template <class T1> void deb(T1 e1) { cout << e1 << endl; } template <class T1, class T2> void deb(T1 e1, T2 e2) { cout << e1 << " " << e2 << endl; } template <class T1, class T2, class T3> void deb(T1 e1, T2 e2, T3 e3) { cout << e1 << " " << e2 << " " << e3 << endl; } void check_time() {} const int res = 1e5 + 5; const long long int mod = 1e9 + 7; int main() { int a, b, c, maxi = 1e9; scanf("%d %d %d", &a, &b, &c); if (a + b > c and a + c > b and b + c > a) { deb("0"); return 0; } if (a + b <= c) { maxi = min(maxi, c - a - b + 1); } if (a + c <= b) { maxi = min(maxi, b - a - c + 1); } if (b + c <= a) { maxi = min(maxi, a - c - b + 1); } deb(maxi); return 0; } ```
#include <bits/stdc++.h> #pragma GCC optimize("Ofast") #pragma GCC target("avx2") #pragma GCC optimization("O3") #pragma GCC optimization("unroll-loops") using namespace std; long long int MOD = 1e9 + 7; long long int ji[300005], ou[300005]; long long int brut = 1e18; int Search_Binary(int arr[], int left, int right, int key) { int midd = 0; while (1) { midd = (left + right) / 2; if (key < arr[midd]) { right = midd - 1; } else if (key > arr[midd]) { left = midd + 1; } else { return midd; } if (left > right) { return -1; } } } int fDig(int i) { int k; do { k = i; i /= 10; } while (i); return k; } int gcd(int a, int b) { while (b) { a %= b; swap(a, b); } return a; } int lcm(int a, int b) { return a / gcd(a, b) * b; } int to_int(string s) { int x; stringstream geek(s); geek >> x; return x; } int prim_const(int x) { vector<bool> prime(x, true); prime[0] = prime[1] = false; for (int i = 2; i * i <= x; ++i) { if (prime[i]) { if (i * 1ll * i <= x) { for (int j = i * i; j <= x; j += i) { prime[j] = false; } } } } if (prime[x] == true) { return 1; } else { return 0; } } int arr[10000]; int key; int main() { cin.tie(0); cout.tie(0); int a, b, c; cin >> a >> b >> c; int minu = 0; if (a + b <= c) { minu += c - a - b + 1; } if (a + c <= b) { minu += b - a - c + 1; } if (b + c <= a) { minu += a - b - c + 1; } cout << minu; cerr << "\nTime elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"; }
### Prompt In cpp, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> #pragma GCC optimize("Ofast") #pragma GCC target("avx2") #pragma GCC optimization("O3") #pragma GCC optimization("unroll-loops") using namespace std; long long int MOD = 1e9 + 7; long long int ji[300005], ou[300005]; long long int brut = 1e18; int Search_Binary(int arr[], int left, int right, int key) { int midd = 0; while (1) { midd = (left + right) / 2; if (key < arr[midd]) { right = midd - 1; } else if (key > arr[midd]) { left = midd + 1; } else { return midd; } if (left > right) { return -1; } } } int fDig(int i) { int k; do { k = i; i /= 10; } while (i); return k; } int gcd(int a, int b) { while (b) { a %= b; swap(a, b); } return a; } int lcm(int a, int b) { return a / gcd(a, b) * b; } int to_int(string s) { int x; stringstream geek(s); geek >> x; return x; } int prim_const(int x) { vector<bool> prime(x, true); prime[0] = prime[1] = false; for (int i = 2; i * i <= x; ++i) { if (prime[i]) { if (i * 1ll * i <= x) { for (int j = i * i; j <= x; j += i) { prime[j] = false; } } } } if (prime[x] == true) { return 1; } else { return 0; } } int arr[10000]; int key; int main() { cin.tie(0); cout.tie(0); int a, b, c; cin >> a >> b >> c; int minu = 0; if (a + b <= c) { minu += c - a - b + 1; } if (a + c <= b) { minu += b - a - c + 1; } if (b + c <= a) { minu += a - b - c + 1; } cout << minu; cerr << "\nTime elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"; } ```
#include <bits/stdc++.h> using namespace std; void solve() { long long a, b, c; cin >> a >> b >> c; long long ans = 0; while (a >= b + c) { b++; ans++; } while (a <= abs(b - c)) { a++; ans++; } cout << ans; } int main() { ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(0); solve(); }
### Prompt In cpp, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; void solve() { long long a, b, c; cin >> a >> b >> c; long long ans = 0; while (a >= b + c) { b++; ans++; } while (a <= abs(b - c)) { a++; ans++; } cout << ans; } int main() { ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(0); solve(); } ```
#include <bits/stdc++.h> using namespace std; int main() { int i, ans, n; n = 3; int ar[n]; for (i = 0; i < n; i++) { cin >> ar[i]; } sort(ar, ar + n); if (ar[0] + ar[1] > ar[2]) { cout << 0 << "\n"; } else { ans = ar[2] - (ar[0] + ar[1]); cout << ans + 1 << "\n"; } return 0; }
### Prompt Your challenge is to write a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int i, ans, n; n = 3; int ar[n]; for (i = 0; i < n; i++) { cin >> ar[i]; } sort(ar, ar + n); if (ar[0] + ar[1] > ar[2]) { cout << 0 << "\n"; } else { ans = ar[2] - (ar[0] + ar[1]); cout << ans + 1 << "\n"; } return 0; } ```
#include <bits/stdc++.h> using namespace std; int32_t main() { ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); long long a[3], b, c; cin >> a[0] >> a[1] >> a[2]; b = a[1]; c = a[2]; if (a[0] + b > c && a[0] + c > b && c + b > a[0]) return cout << 0, 0; sort(a, a + 3); long long ans = 0; b = a[0]; c = a[1]; while (1) { b++; ans++; if (b + c > a[2]) return cout << ans, 0; c++; ans++; if (b + c > a[2]) return cout << ans, 0; } }
### Prompt Your task is to create a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int32_t main() { ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); long long a[3], b, c; cin >> a[0] >> a[1] >> a[2]; b = a[1]; c = a[2]; if (a[0] + b > c && a[0] + c > b && c + b > a[0]) return cout << 0, 0; sort(a, a + 3); long long ans = 0; b = a[0]; c = a[1]; while (1) { b++; ans++; if (b + c > a[2]) return cout << ans, 0; c++; ans++; if (b + c > a[2]) return cout << ans, 0; } } ```
#include <bits/stdc++.h> int main() { int a, b, c, x, y, z, dif; scanf("%d", &a); scanf("%d", &b); scanf("%d", &c); if (a >= b && a >= c) { x = a; y = b; z = c; } else if (b >= c && b >= a) { x = b; y = a; z = c; } else if (c >= b && c >= a) { x = c; y = b; z = a; } if (y + z <= x) { dif = x - y - z; printf("%d", dif + 1); } else printf("0"); }
### Prompt Your challenge is to write a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> int main() { int a, b, c, x, y, z, dif; scanf("%d", &a); scanf("%d", &b); scanf("%d", &c); if (a >= b && a >= c) { x = a; y = b; z = c; } else if (b >= c && b >= a) { x = b; y = a; z = c; } else if (c >= b && c >= a) { x = c; y = b; z = a; } if (y + z <= x) { dif = x - y - z; printf("%d", dif + 1); } else printf("0"); } ```
#include <bits/stdc++.h> using namespace std; const int MAXN = 100005; int arr[MAXN]; int main() { ios::sync_with_stdio(0); cin.tie(0); cin >> arr[0] >> arr[1] >> arr[2]; sort(arr, arr + 3); cout << max(0, arr[2] + 1 - (arr[0] + arr[1])) << "\n"; return 0; }
### Prompt Your task is to create a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = 100005; int arr[MAXN]; int main() { ios::sync_with_stdio(0); cin.tie(0); cin >> arr[0] >> arr[1] >> arr[2]; sort(arr, arr + 3); cout << max(0, arr[2] + 1 - (arr[0] + arr[1])) << "\n"; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int sides[3], addTime; for (int i = 0; i < 3; i++) { cin >> sides[i]; } sort(sides, sides + 3); if (sides[0] + sides[1] > sides[2]) { cout << "0" << endl; } else { addTime = (sides[2] + 1) - (sides[0] + sides[1]); cout << addTime << endl; } return 0; }
### Prompt Please provide a CPP coded solution to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int sides[3], addTime; for (int i = 0; i < 3; i++) { cin >> sides[i]; } sort(sides, sides + 3); if (sides[0] + sides[1] > sides[2]) { cout << "0" << endl; } else { addTime = (sides[2] + 1) - (sides[0] + sides[1]); cout << addTime << endl; } return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c; cin >> a >> b >> c; if (a < b) swap(a, b); if (b < c) swap(b, c); cout << max(a + 1 - b - c, max(b + 1 - a - c, max(c + 1 - b - a, 0))) << endl; return 0; }
### Prompt Your task is to create a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c; cin >> a >> b >> c; if (a < b) swap(a, b); if (b < c) swap(b, c); cout << max(a + 1 - b - c, max(b + 1 - a - c, max(c + 1 - b - a, 0))) << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c; while (~scanf("%d%d%d", &a, &b, &c)) { int a1 = b + c, b1 = a + c, c1 = a + b; if (a1 > a && b1 > b && c1 > c) printf("0\n"); else { int minn = 9999999; minn = minn > (a1 - a) ? (a1 - a) : minn; minn = minn > (b1 - b) ? (b1 - b) : minn; minn = minn > (c1 - c) ? (c1 - c) : minn; printf("%d\n", -minn + 1); } } return 0; }
### Prompt Create a solution in cpp for the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c; while (~scanf("%d%d%d", &a, &b, &c)) { int a1 = b + c, b1 = a + c, c1 = a + b; if (a1 > a && b1 > b && c1 > c) printf("0\n"); else { int minn = 9999999; minn = minn > (a1 - a) ? (a1 - a) : minn; minn = minn > (b1 - b) ? (b1 - b) : minn; minn = minn > (c1 - c) ? (c1 - c) : minn; printf("%d\n", -minn + 1); } } return 0; } ```
#include <bits/stdc++.h> using namespace std; const int MX = 100005, infi = 1000000000, mod = 1000000007; int n, v[3]; int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); cin >> v[0] >> v[1] >> v[2]; sort(v, v + 3); cout << max(0, v[2] - (v[0] + v[1]) + 1); return 0; }
### Prompt Create a solution in Cpp for the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; const int MX = 100005, infi = 1000000000, mod = 1000000007; int n, v[3]; int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); cin >> v[0] >> v[1] >> v[2]; sort(v, v + 3); cout << max(0, v[2] - (v[0] + v[1]) + 1); return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); long long i, arr[3]; for (i = 0; i < 3; i++) { cin >> arr[i]; } sort(arr, arr + 3); if ((arr[0] + arr[1]) > arr[2]) cout << "0"; else cout << (arr[2] - (arr[0] + arr[1])) + 1; return 0; }
### Prompt Please provide a CPP coded solution to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); long long i, arr[3]; for (i = 0; i < 3; i++) { cin >> arr[i]; } sort(arr, arr + 3); if ((arr[0] + arr[1]) > arr[2]) cout << "0"; else cout << (arr[2] - (arr[0] + arr[1])) + 1; return 0; } ```
#include <bits/stdc++.h> #pragma GCC optimize("O3") using namespace std; void io() {} long long fast_exp(long long base, int exp, int m) { long long res = 1; while (exp > 0) { if (exp % 2 == 1) { res = (res * base) % m; } base = (base * base) % m; exp /= 2; } return res % m; } long long arr[200005]; long long val[200005]; long long siz[200005]; map<long long, long long> mp; long long ans, cnt, tmp; long long vis[200005]; vector<long long> adj[200005]; long long dp[2001][2001]; long long n, m, k; int main() { io(); ios ::sync_with_stdio(false); cin.tie(0); cout.tie(0); long long t = 0; cin >> arr[0] >> arr[1] >> arr[2]; sort(arr, arr + 3); if (arr[2] - arr[1] - arr[0] < 0) { cout << 0 << '\n'; return 0; } cout << arr[2] - arr[1] - arr[0] + 1 << '\n'; }
### Prompt Generate a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> #pragma GCC optimize("O3") using namespace std; void io() {} long long fast_exp(long long base, int exp, int m) { long long res = 1; while (exp > 0) { if (exp % 2 == 1) { res = (res * base) % m; } base = (base * base) % m; exp /= 2; } return res % m; } long long arr[200005]; long long val[200005]; long long siz[200005]; map<long long, long long> mp; long long ans, cnt, tmp; long long vis[200005]; vector<long long> adj[200005]; long long dp[2001][2001]; long long n, m, k; int main() { io(); ios ::sync_with_stdio(false); cin.tie(0); cout.tie(0); long long t = 0; cin >> arr[0] >> arr[1] >> arr[2]; sort(arr, arr + 3); if (arr[2] - arr[1] - arr[0] < 0) { cout << 0 << '\n'; return 0; } cout << arr[2] - arr[1] - arr[0] + 1 << '\n'; } ```
#include <bits/stdc++.h> using namespace std; void solve() { int a[5]; cin >> a[1] >> a[2] >> a[3]; sort(a + 1, a + 3 + 1); if (a[1] + a[2] > a[3]) cout << 0 << endl; else cout << a[3] + 1 - a[1] - a[2] << endl; } int main() { solve(); return 0; }
### Prompt Generate a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; void solve() { int a[5]; cin >> a[1] >> a[2] >> a[3]; sort(a + 1, a + 3 + 1); if (a[1] + a[2] > a[3]) cout << 0 << endl; else cout << a[3] + 1 - a[1] - a[2] << endl; } int main() { solve(); return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); if (a[0] + a[1] > a[2]) cout << 0; else if (a[0] + a[1] == a[2]) cout << 1; else cout << (a[2] - (a[0] + a[1])) + 1; return 0; }
### Prompt Your task is to create a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); if (a[0] + a[1] > a[2]) cout << 0; else if (a[0] + a[1] == a[2]) cout << 1; else cout << (a[2] - (a[0] + a[1])) + 1; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; int n = sizeof(a) / sizeof(a[0]); sort(a, a + n); if (a[0] + a[1] > a[2]) { cout << "0"; } else if (a[0] + a[1] <= a[2]) { cout << a[2] - (a[0] + a[1]) + 1; } }
### Prompt Please formulate a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; int n = sizeof(a) / sizeof(a[0]); sort(a, a + n); if (a[0] + a[1] > a[2]) { cout << "0"; } else if (a[0] + a[1] <= a[2]) { cout << a[2] - (a[0] + a[1]) + 1; } } ```
#include <bits/stdc++.h> using namespace std; int main() { int arr[3]; for (int i = 0; i < 3; i++) { cin >> arr[i]; } sort(arr, arr + 3); int val = arr[0] + arr[1]; if (val <= arr[2]) { int res = arr[2] - val + 1; cout << res << endl; } else cout << "0" << endl; }
### Prompt Develop a solution in Cpp to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int arr[3]; for (int i = 0; i < 3; i++) { cin >> arr[i]; } sort(arr, arr + 3); int val = arr[0] + arr[1]; if (val <= arr[2]) { int res = arr[2] - val + 1; cout << res << endl; } else cout << "0" << endl; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[3]; for (int i = 0; i < 3; i++) cin >> a[i]; sort(a, a + 3); int y = a[2] + 1 - a[0] - a[1]; cout << max(0, y); return 0; }
### Prompt In Cpp, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[3]; for (int i = 0; i < 3; i++) cin >> a[i]; sort(a, a + 3); int y = a[2] + 1 - a[0] - a[1]; cout << max(0, y); return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c; cin >> a >> b >> c; if (a < b) swap(a, b); if (a < c) swap(a, c); if (b < c) swap(b, c); if (b + c <= a) cout << a - c - b + 1; else cout << 0; }
### Prompt Construct a CPP code solution to the problem outlined: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c; cin >> a >> b >> c; if (a < b) swap(a, b); if (a < c) swap(a, c); if (b < c) swap(b, c); if (b + c <= a) cout << a - c - b + 1; else cout << 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c; scanf("%d %d %d", &a, &b, &c); if (a < b) swap(a, b); if (a < c) swap(a, c); int ans; if (b + c > a) ans = 0; else ans = a - b - c + 1; printf("%d\n", ans); return 0; }
### Prompt Construct a CPP code solution to the problem outlined: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c; scanf("%d %d %d", &a, &b, &c); if (a < b) swap(a, b); if (a < c) swap(a, c); int ans; if (b + c > a) ans = 0; else ans = a - b - c + 1; printf("%d\n", ans); return 0; } ```
#include <bits/stdc++.h> using namespace std; inline int mul(int a, int b) { return int(a * 1ll * b % 998244353); } inline int norm(int a) { if (a >= 998244353) a -= 998244353; if (a < 0) a += 998244353; return a; } inline int binPow(int a, int k) { int ans = 1; while (k > 0) { if (k & 1) ans = mul(ans, a); a = mul(a, a); k >>= 1; } return ans; } inline int subtract(int a, int b) { return norm(a - b); } inline int sum(int a, int b) { return int((a + b) % 998244353); } inline int inv(int a) { return binPow(a, 998244353 - 2); } int main() { cin.sync_with_stdio(false); cin.tie(0); vector<int> a(3); for (int i = 0; i < 3; i++) { cin >> a[i]; } sort(a.begin(), a.end()); cout << max(0, a[2] - a[1] - a[0] + 1); return 0; }
### Prompt Generate a cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; inline int mul(int a, int b) { return int(a * 1ll * b % 998244353); } inline int norm(int a) { if (a >= 998244353) a -= 998244353; if (a < 0) a += 998244353; return a; } inline int binPow(int a, int k) { int ans = 1; while (k > 0) { if (k & 1) ans = mul(ans, a); a = mul(a, a); k >>= 1; } return ans; } inline int subtract(int a, int b) { return norm(a - b); } inline int sum(int a, int b) { return int((a + b) % 998244353); } inline int inv(int a) { return binPow(a, 998244353 - 2); } int main() { cin.sync_with_stdio(false); cin.tie(0); vector<int> a(3); for (int i = 0; i < 3; i++) { cin >> a[i]; } sort(a.begin(), a.end()); cout << max(0, a[2] - a[1] - a[0] + 1); return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int arr[3], r = 0; cin >> arr[0] >> arr[1] >> arr[2]; sort(arr, arr + 3); if (arr[0] + arr[1] > arr[2]) { cout << "0"; } else { cout << arr[2] - (arr[0] + arr[1]) + 1; } }
### Prompt Your task is to create a cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int arr[3], r = 0; cin >> arr[0] >> arr[1] >> arr[2]; sort(arr, arr + 3); if (arr[0] + arr[1] > arr[2]) { cout << "0"; } else { cout << arr[2] - (arr[0] + arr[1]) + 1; } } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c, ctr = 0; cin >> a >> b >> c; while (a + b <= c) { ctr++; a > b ? a++ : b++; } while (b + c <= a) { ctr++; b > c ? b++ : c++; } while (a + c <= b) { ctr++; a > c ? a++ : c++; } cout << ctr << "\n"; }
### Prompt Please provide a CPP coded solution to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c, ctr = 0; cin >> a >> b >> c; while (a + b <= c) { ctr++; a > b ? a++ : b++; } while (b + c <= a) { ctr++; b > c ? b++ : c++; } while (a + c <= b) { ctr++; a > c ? a++ : c++; } cout << ctr << "\n"; } ```
#include <bits/stdc++.h> using namespace std; int main(void) { int a[3]; for (int i = 0; i < 3; i++) cin >> a[i]; sort(a, a + 3); int sum = 0; for (int i = 0;; i++) { if (a[1] + a[0] > a[2]) break; else { sum++; a[1] += 1; } } cout << sum; return 0; }
### Prompt Your task is to create a cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main(void) { int a[3]; for (int i = 0; i < 3; i++) cin >> a[i]; sort(a, a + 3); int sum = 0; for (int i = 0;; i++) { if (a[1] + a[0] > a[2]) break; else { sum++; a[1] += 1; } } cout << sum; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int arr[3]; cin >> arr[0] >> arr[1] >> arr[2]; sort(arr, arr + 3); if (arr[0] + arr[1] > arr[2]) { cout << 0; } else { cout << arr[2] - arr[1] - arr[0] + 1; } }
### Prompt Construct a cpp code solution to the problem outlined: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int arr[3]; cin >> arr[0] >> arr[1] >> arr[2]; sort(arr, arr + 3); if (arr[0] + arr[1] > arr[2]) { cout << 0; } else { cout << arr[2] - arr[1] - arr[0] + 1; } } ```
#include <bits/stdc++.h> using namespace std; int main() { int arr[] = {0, 0, 0}; for (int i = 0; i < 3; i++) cin >> arr[i]; sort(arr, arr + 3); if (arr[2] < (arr[0] + arr[1])) cout << "0"; else cout << (1 + (arr[2] - (arr[0] + arr[1]))); return 0; }
### Prompt In CPP, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int arr[] = {0, 0, 0}; for (int i = 0; i < 3; i++) cin >> arr[i]; sort(arr, arr + 3); if (arr[2] < (arr[0] + arr[1])) cout << "0"; else cout << (1 + (arr[2] - (arr[0] + arr[1]))); return 0; } ```
#include <bits/stdc++.h> int main() { int a[10], b, c, i, j; for (i = 0; i < 3; i++) scanf("%d", &a[i]); for (i = 0; i < 3; i++) { for (j = i + 1; j < 3; j++) { if (a[i] > a[j]) { b = a[i]; a[i] = a[j]; a[j] = b; } } } int p = a[0], q = a[1], r = a[2], z; if ((p + q) <= r) { z = r - (p + q) + 1; printf("%d\n", z); } else printf("0\n"); }
### Prompt Develop a solution in Cpp to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> int main() { int a[10], b, c, i, j; for (i = 0; i < 3; i++) scanf("%d", &a[i]); for (i = 0; i < 3; i++) { for (j = i + 1; j < 3; j++) { if (a[i] > a[j]) { b = a[i]; a[i] = a[j]; a[j] = b; } } } int p = a[0], q = a[1], r = a[2], z; if ((p + q) <= r) { z = r - (p + q) + 1; printf("%d\n", z); } else printf("0\n"); } ```
#include <bits/stdc++.h> using namespace std; int a, b, c, A, B, C; string s; int main() { cin >> a >> b >> c; A = max(a, b); B = max(b, c); C = max(A, B); if (C == a) { if (a < b + c) { cout << 0; return 0; } else { cout << a - b - c + 1; return 0; } } if (C == b) { if (b < a + c) { cout << 0; return 0; } else { cout << b - a - c + 1; return 0; } } if (C == c) { if (c < a + b) { cout << 0; return 0; } else { cout << c - a - b + 1; return 0; } } }
### Prompt Please formulate a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int a, b, c, A, B, C; string s; int main() { cin >> a >> b >> c; A = max(a, b); B = max(b, c); C = max(A, B); if (C == a) { if (a < b + c) { cout << 0; return 0; } else { cout << a - b - c + 1; return 0; } } if (C == b) { if (b < a + c) { cout << 0; return 0; } else { cout << b - a - c + 1; return 0; } } if (C == c) { if (c < a + b) { cout << 0; return 0; } else { cout << c - a - b + 1; return 0; } } } ```
#include <bits/stdc++.h> using namespace std; const int inf = 1e9 + 3; const long long Linf = 1e18; int main() { ios_base::sync_with_stdio(false); int x[3]; cin >> x[0] >> x[1] >> x[2]; sort(x, x + 3); int ans = (x[2] >= x[0] + x[1] ? x[2] - x[1] - x[0] + 1 : 0); cout << ans << '\n'; return 0; }
### Prompt Construct a cpp code solution to the problem outlined: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; const int inf = 1e9 + 3; const long long Linf = 1e18; int main() { ios_base::sync_with_stdio(false); int x[3]; cin >> x[0] >> x[1] >> x[2]; sort(x, x + 3); int ans = (x[2] >= x[0] + x[1] ? x[2] - x[1] - x[0] + 1 : 0); cout << ans << '\n'; return 0; } ```
#include <bits/stdc++.h> using namespace std; int a, b, c, Count = 1; int main() { cin >> a >> b >> c; if ((a + b > c) && (a + c > b) && (b + c > a)) { cout << 0 << endl; return 0; } while ((a + b < c) || (a + c < b) || (b + c < a)) { ++Count; if (a + b < c) { if (a < b) a++; else b++; } else if (a + c < b) { if (a < c) a++; else c++; } else if (b + c < a) { if (b < c) b++; else c++; } } cout << Count << endl; return 0; }
### Prompt Your task is to create a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int a, b, c, Count = 1; int main() { cin >> a >> b >> c; if ((a + b > c) && (a + c > b) && (b + c > a)) { cout << 0 << endl; return 0; } while ((a + b < c) || (a + c < b) || (b + c < a)) { ++Count; if (a + b < c) { if (a < b) a++; else b++; } else if (a + c < b) { if (a < c) a++; else c++; } else if (b + c < a) { if (b < c) b++; else c++; } } cout << Count << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; int n[3], cnt; int main() { for (int i = 0; i < 3; i++) { cin >> n[i]; } sort(n, n + 3); if (n[0] + n[1] > n[2]) { cout << 0; return 0; } else { while (n[0] + n[1] <= n[2]) { n[1]++; cnt++; } } cout << cnt; }
### Prompt In CPP, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int n[3], cnt; int main() { for (int i = 0; i < 3; i++) { cin >> n[i]; } sort(n, n + 3); if (n[0] + n[1] > n[2]) { cout << 0; return 0; } else { while (n[0] + n[1] <= n[2]) { n[1]++; cnt++; } } cout << cnt; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); int cnt = 0; while (a[0] + a[1] <= a[2]) { cnt++, a[0]++; } cout << cnt << endl; return 0; }
### Prompt Construct a cpp code solution to the problem outlined: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); int cnt = 0; while (a[0] + a[1] <= a[2]) { cnt++, a[0]++; } cout << cnt << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c, d, e, f, i; cin >> a >> b >> c; d = a + b; e = b + c; f = a + c; int sum1 = 0, sum2 = 0, sum3 = 0; for (i = 0; i < 100; i++) { if (d > c) break; else { sum1++; d++; } } for (i = 0; i < 100; i++) { if (e > a) break; else { sum2++; e++; } } for (i = 0; i < 100; i++) { if (f > b) break; else { sum3++; f++; } } int res = sum1 + sum2 + sum3; cout << res << endl; return 0; }
### Prompt Your task is to create a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c, d, e, f, i; cin >> a >> b >> c; d = a + b; e = b + c; f = a + c; int sum1 = 0, sum2 = 0, sum3 = 0; for (i = 0; i < 100; i++) { if (d > c) break; else { sum1++; d++; } } for (i = 0; i < 100; i++) { if (e > a) break; else { sum2++; e++; } } for (i = 0; i < 100; i++) { if (f > b) break; else { sum3++; f++; } } int res = sum1 + sum2 + sum3; cout << res << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c; int ans = 0; cin >> a >> b >> c; while (a + b <= c || b + c <= a || a + c <= b) { if (a + b <= c) { if (a < b) { a++; } else { b++; } ans++; } if (b + c <= a) { if (b < c) { b++; } else { c++; } ans++; } if (a + c <= b) { if (a < c) { a++; } else { c++; } ans++; } } cout << ans; return 0; }
### Prompt Create a solution in cpp for the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c; int ans = 0; cin >> a >> b >> c; while (a + b <= c || b + c <= a || a + c <= b) { if (a + b <= c) { if (a < b) { a++; } else { b++; } ans++; } if (b + c <= a) { if (b < c) { b++; } else { c++; } ans++; } if (a + c <= b) { if (a < c) { a++; } else { c++; } ans++; } } cout << ans; return 0; } ```
#include <bits/stdc++.h> using namespace std; vector<int> v; int main() { int a, b, c; cin >> a >> b >> c; if ((a + b > c) && (b + c > a) && (c + a > b)) cout << "0\n"; else { v.push_back(a); v.push_back(b); v.push_back(c); sort(v.begin(), v.end()); a = v[0]; b = v[1]; c = v[2]; cout << (c - (a + b) + 1) << "\n"; } }
### Prompt In CPP, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; vector<int> v; int main() { int a, b, c; cin >> a >> b >> c; if ((a + b > c) && (b + c > a) && (c + a > b)) cout << "0\n"; else { v.push_back(a); v.push_back(b); v.push_back(c); sort(v.begin(), v.end()); a = v[0]; b = v[1]; c = v[2]; cout << (c - (a + b) + 1) << "\n"; } } ```
#include <bits/stdc++.h> #pragma GCC optimize("O3") std::pair<int, int> DR[] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}, {-1, 1}, {-1, -1}, {1, 1}, {1, -1}}; using namespace std; int a[3]; int32_t main() { ios_base::sync_with_stdio(false); cin.tie(0); cerr.tie(0); cout.tie(0); cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); return cout << max(0, a[2] - a[0] - a[1] + 1), 0; }
### Prompt Generate a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> #pragma GCC optimize("O3") std::pair<int, int> DR[] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}, {-1, 1}, {-1, -1}, {1, 1}, {1, -1}}; using namespace std; int a[3]; int32_t main() { ios_base::sync_with_stdio(false); cin.tie(0); cerr.tie(0); cout.tie(0); cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); return cout << max(0, a[2] - a[0] - a[1] + 1), 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); if (a[0] + a[1] > a[2]) cout << 0; else cout << a[2] - a[0] - a[1] + 1; return 0; }
### Prompt Construct a cpp code solution to the problem outlined: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); if (a[0] + a[1] > a[2]) cout << 0; else cout << a[2] - a[0] - a[1] + 1; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[3]; for (int i = 0; i < 3; i++) { cin >> a[i]; } sort(a, a + 3); if (a[2] < a[0] + a[1]) { cout << 0 << endl; } else { cout << a[2] - a[1] - a[0] + 1 << endl; } return 0; }
### Prompt Generate a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[3]; for (int i = 0; i < 3; i++) { cin >> a[i]; } sort(a, a + 3); if (a[2] < a[0] + a[1]) { cout << 0 << endl; } else { cout << a[2] - a[1] - a[0] + 1 << endl; } return 0; } ```
#include <bits/stdc++.h> using namespace std; long long a[3]; int main() { ios_base::sync_with_stdio(false); cin.tie(0); cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); if (a[0] + a[1] > a[2]) { cout << 0; return 0; } cout << a[2] - (a[1] + a[0]) + 1; return 0; }
### Prompt Please create a solution in Cpp to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; long long a[3]; int main() { ios_base::sync_with_stdio(false); cin.tie(0); cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); if (a[0] + a[1] > a[2]) { cout << 0; return 0; } cout << a[2] - (a[1] + a[0]) + 1; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int T = 1; while (T--) { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); cout << max(0, a[2] - a[1] - a[0] + 1); } return 0; }
### Prompt In cpp, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int T = 1; while (T--) { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); cout << max(0, a[2] - a[1] - a[0] + 1); } return 0; } ```
#include <bits/stdc++.h> using namespace std; long long ans; int main() { vector<int> a(3); for (size_t i = 0; i < 3; i++) { cin >> a[i]; } sort(a.begin(), a.end()); cout << ((a[1] + a[0] > a[2]) ? 0 : (-a[1] - a[0] + a[2] + 1)); }
### Prompt Your challenge is to write a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; long long ans; int main() { vector<int> a(3); for (size_t i = 0; i < 3; i++) { cin >> a[i]; } sort(a.begin(), a.end()); cout << ((a[1] + a[0] > a[2]) ? 0 : (-a[1] - a[0] + a[2] + 1)); } ```
#include <bits/stdc++.h> using namespace std; bool checkValidity(int a, int b, int c) { if (a + b <= c || a + c <= b || b + c <= a) return false; else return true; } int maximum(int a, int b, int c) { int max = (a < b) ? b : a; return ((max < c) ? c : max); } int main() { long long a, b, c, d; cin >> a >> b >> c; if (checkValidity(a, b, c)) cout << 0; else { d = maximum(a, b, c); if (a == d) cout << (a - (b + c) + 1); else if (b == d) cout << (b - (a + c) + 1); else cout << (c - (a + b) + 1); } return 0; }
### Prompt Your challenge is to write a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; bool checkValidity(int a, int b, int c) { if (a + b <= c || a + c <= b || b + c <= a) return false; else return true; } int maximum(int a, int b, int c) { int max = (a < b) ? b : a; return ((max < c) ? c : max); } int main() { long long a, b, c, d; cin >> a >> b >> c; if (checkValidity(a, b, c)) cout << 0; else { d = maximum(a, b, c); if (a == d) cout << (a - (b + c) + 1); else if (b == d) cout << (b - (a + c) + 1); else cout << (c - (a + b) + 1); } return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c, ans; cin >> a >> b >> c; int a1[3]; a1[0] = a; a1[1] = b; a1[2] = c; sort(a1, a1 + 3); if (a1[0] + a1[1] > a1[2]) ans = 0; else ans = a1[2] - a1[0] - a1[1] + 1; cout << ans; return 0; }
### Prompt Please formulate a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c, ans; cin >> a >> b >> c; int a1[3]; a1[0] = a; a1[1] = b; a1[2] = c; sort(a1, a1 + 3); if (a1[0] + a1[1] > a1[2]) ans = 0; else ans = a1[2] - a1[0] - a1[1] + 1; cout << ans; return 0; } ```
#include <bits/stdc++.h> using namespace std; int code() { long long int a, b, c; cin >> a >> b >> c; long long int aa = max(a, b); aa = max(aa, c); long long int sum = a + b + c - aa; long long int ans = ((sum - aa <= 0) ? aa - sum + 1 : 0); cout << ans << endl; return 0; } signed main() { ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL); code(); }
### Prompt Create a solution in cpp for the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int code() { long long int a, b, c; cin >> a >> b >> c; long long int aa = max(a, b); aa = max(aa, c); long long int sum = a + b + c - aa; long long int ans = ((sum - aa <= 0) ? aa - sum + 1 : 0); cout << ans << endl; return 0; } signed main() { ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL); code(); } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c, x, y, z; cin >> a >> b >> c; int q[3] = {a, b, c}; sort(q, q + 3, greater<int>()); x = q[0]; y = q[1]; z = q[2]; int t = z + y; if (t > x) cout << 0; else { cout << (x - t) + 1; } }
### Prompt Your task is to create a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c, x, y, z; cin >> a >> b >> c; int q[3] = {a, b, c}; sort(q, q + 3, greater<int>()); x = q[0]; y = q[1]; z = q[2]; int t = z + y; if (t > x) cout << 0; else { cout << (x - t) + 1; } } ```
#include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(0); cin.tie(0); vector<int> v(3); cin >> v[0] >> v[1] >> v[2]; sort(v.begin(), v.end()); int ans = max(0, v[2] - v[1] - v[0] + 1); cout << ans << endl; return 0; }
### Prompt Develop a solution in Cpp to the problem described below: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(0); cin.tie(0); vector<int> v(3); cin >> v[0] >> v[1] >> v[2]; sort(v.begin(), v.end()); int ans = max(0, v[2] - v[1] - v[0] + 1); cout << ans << endl; return 0; } ```
#include <bits/stdc++.h> int main() { unsigned int x[3], i, t, n = 0; for (i = 0; i < 3; i++) scanf("%u", &x[i]); if (x[2] < x[1]) { t = x[1]; x[1] = x[2]; x[2] = t; } if (x[2] < x[0]) { t = x[0]; x[0] = x[2]; x[2] = t; } while (!(x[0] + x[1] > x[2])) { x[2]--; n++; } printf("%u", n); return 0; }
### Prompt Your task is to create a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> int main() { unsigned int x[3], i, t, n = 0; for (i = 0; i < 3; i++) scanf("%u", &x[i]); if (x[2] < x[1]) { t = x[1]; x[1] = x[2]; x[2] = t; } if (x[2] < x[0]) { t = x[0]; x[0] = x[2]; x[2] = t; } while (!(x[0] + x[1] > x[2])) { x[2]--; n++; } printf("%u", n); return 0; } ```
#include <bits/stdc++.h> using namespace std; int main(int argc, char *argv[]) { vector<int> Length(3); for (int i = 0; i < 3; ++i) { scanf("%d", &Length[i]); } sort(Length.begin(), Length.end()); printf("%d\n", max(0, Length[2] - (Length[0] + Length[1]) + 1)); return 0; }
### Prompt Create a solution in Cpp for the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main(int argc, char *argv[]) { vector<int> Length(3); for (int i = 0; i < 3; ++i) { scanf("%d", &Length[i]); } sort(Length.begin(), Length.end()); printf("%d\n", max(0, Length[2] - (Length[0] + Length[1]) + 1)); return 0; } ```
#include <bits/stdc++.h> using namespace std; long long int power(long long int x, long long int y, long long int p) { long long int res = 1; x = x % p; while (y > 0) { if (y & 1) res = (res * x) % p; y = y / 2; x = (x * x) % p; } return res; } void solve() { long long int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); long long int x = a[0] + a[1] - a[2]; if (x > 0) { cout << 0 << "\n"; return; } else { cout << a[2] + 1 - (a[0] + a[1]) << "\n"; return; } } int main() { long long int t = 1; while (t--) { solve(); } return 0; }
### Prompt Construct a Cpp code solution to the problem outlined: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; long long int power(long long int x, long long int y, long long int p) { long long int res = 1; x = x % p; while (y > 0) { if (y & 1) res = (res * x) % p; y = y / 2; x = (x * x) % p; } return res; } void solve() { long long int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); long long int x = a[0] + a[1] - a[2]; if (x > 0) { cout << 0 << "\n"; return; } else { cout << a[2] + 1 - (a[0] + a[1]) << "\n"; return; } } int main() { long long int t = 1; while (t--) { solve(); } return 0; } ```
#include <bits/stdc++.h> int main() { int a, b, c, x = 0; scanf("%d%d%d", &a, &b, &c); if ((a + b) <= c) { x = x + (c - (a + b)) + 1; } if ((b + c) <= a) { x = x + (a - (b + c)) + 1; } if ((a + c) <= b) { x = x + (b - (a + c)) + 1; } printf("%d", x); return 0; }
### Prompt Create a solution in CPP for the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> int main() { int a, b, c, x = 0; scanf("%d%d%d", &a, &b, &c); if ((a + b) <= c) { x = x + (c - (a + b)) + 1; } if ((b + c) <= a) { x = x + (a - (b + c)) + 1; } if ((a + c) <= b) { x = x + (b - (a + c)) + 1; } printf("%d", x); return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); long long a, b, c; cin >> a >> b >> c; long long fans = 0; if (c >= a && c >= b) { long long ans = c - a - b; if (ans < 0) { fans = 0; } else { fans = ans + 1; } } else if (b >= a && b >= c) { long long ans = b - a - c; if (ans < 0) { fans = 0; } else { fans = ans + 1; } } else { long long ans = a - b - c; if (ans < 0) { fans = 0; } else { fans = ans + 1; } } cout << fans << endl; return 0; }
### Prompt Construct a cpp code solution to the problem outlined: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); long long a, b, c; cin >> a >> b >> c; long long fans = 0; if (c >= a && c >= b) { long long ans = c - a - b; if (ans < 0) { fans = 0; } else { fans = ans + 1; } } else if (b >= a && b >= c) { long long ans = b - a - c; if (ans < 0) { fans = 0; } else { fans = ans + 1; } } else { long long ans = a - b - c; if (ans < 0) { fans = 0; } else { fans = ans + 1; } } cout << fans << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(0); int a, b, c; cin >> a >> b >> c; int inf = 0x3f3f3f3f; int ans = inf; for (int i = a; i <= 100; i++) { for (int j = b; j <= 100; j++) { for (int k = c; k <= 100; k++) { if (i + j > k && i + k > j && j + k > i) ans = min(ans, i - a + j - b + k - c); } } } cout << ans; return 0; }
### Prompt In cpp, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(0); int a, b, c; cin >> a >> b >> c; int inf = 0x3f3f3f3f; int ans = inf; for (int i = a; i <= 100; i++) { for (int j = b; j <= 100; j++) { for (int k = c; k <= 100; k++) { if (i + j > k && i + k > j && j + k > i) ans = min(ans, i - a + j - b + k - c); } } } cout << ans; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); long long int a[3], i; for (i = 0; i < 3; i++) cin >> a[i]; sort(a, a + 3); long long int k = a[2] - (a[1] + a[0]); if (k < 0) cout << 0 << endl; else cout << k + 1 << endl; }
### Prompt Please formulate a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); long long int a[3], i; for (i = 0; i < 3; i++) cin >> a[i]; sort(a, a + 3); long long int k = a[2] - (a[1] + a[0]); if (k < 0) cout << 0 << endl; else cout << k + 1 << endl; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c, d = 0; cin >> a >> b >> c; if (a >= b + c) { d = a - (b + c); d++; } else if (b >= a + c) { d = b - (a + c); d++; } else if (c >= a + b) { d = c - (a + b); d++; } cout << d; return 0; }
### Prompt Your challenge is to write a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c, d = 0; cin >> a >> b >> c; if (a >= b + c) { d = a - (b + c); d++; } else if (b >= a + c) { d = b - (a + c); d++; } else if (c >= a + b) { d = c - (a + b); d++; } cout << d; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int a, b, c, ans = 0; cin >> a >> b >> c; if (c >= a && c >= b) ans = max(0, c + 1 - (a + b)); else if (b >= a && b >= c) ans = max(0, b + 1 - (a + c)); else if (a >= b && a >= c) ans = max(0, a + 1 - (b + c)); cout << ans << endl; return 0; }
### Prompt In CPP, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int a, b, c, ans = 0; cin >> a >> b >> c; if (c >= a && c >= b) ans = max(0, c + 1 - (a + b)); else if (b >= a && b >= c) ans = max(0, b + 1 - (a + c)); else if (a >= b && a >= c) ans = max(0, a + 1 - (b + c)); cout << ans << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c, d, e; cin >> a >> b >> c; d = max(c, max(a, b)); if (d == a) { if (b + c >= d + 1) { cout << "0" << endl; } else { e = (d + 1) - (b + c); cout << e << endl; } } else if (d == b) { if (a + c >= d + 1) { cout << "0" << endl; } else { e = (d + 1) - (a + c); cout << e << endl; } } else if (d == c) { if (a + b >= d + 1) cout << "0" << endl; else { e = (d + 1) - (a + b); cout << e << endl; } } return 0; }
### Prompt Your challenge is to write a Cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c, d, e; cin >> a >> b >> c; d = max(c, max(a, b)); if (d == a) { if (b + c >= d + 1) { cout << "0" << endl; } else { e = (d + 1) - (b + c); cout << e << endl; } } else if (d == b) { if (a + c >= d + 1) { cout << "0" << endl; } else { e = (d + 1) - (a + c); cout << e << endl; } } else if (d == c) { if (a + b >= d + 1) cout << "0" << endl; else { e = (d + 1) - (a + b); cout << e << endl; } } return 0; } ```
#include <bits/stdc++.h> using namespace std; void rewenie(int b) { int c = b / 7; if (b % 7 != 0) c++; cout << c; } int main() { int a, b, c; cin >> a >> b >> c; int sum = 0; if (a >= b && a >= c) { if (a >= b + c) cout << (a - b - c) + 1; else goto A; } else if (a <= b && b >= c) { if (b >= a + c) cout << (b - a - c) + 1; else goto A; } else if (a <= c && b <= c) { if (c >= a + b) cout << (c - a - b) + 1; else goto A; } else { A: cout << "0"; } }
### Prompt Construct a Cpp code solution to the problem outlined: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; void rewenie(int b) { int c = b / 7; if (b % 7 != 0) c++; cout << c; } int main() { int a, b, c; cin >> a >> b >> c; int sum = 0; if (a >= b && a >= c) { if (a >= b + c) cout << (a - b - c) + 1; else goto A; } else if (a <= b && b >= c) { if (b >= a + c) cout << (b - a - c) + 1; else goto A; } else if (a <= c && b <= c) { if (c >= a + b) cout << (c - a - b) + 1; else goto A; } else { A: cout << "0"; } } ```
#include <bits/stdc++.h> using namespace std; int main() { int sum = 0, maxi = 0, x; for (int i = 0; i < 3; i++) { cin >> x; if (x > maxi) { maxi = x; } sum += x; } if (sum - maxi <= maxi) { cout << maxi - (sum - maxi) + 1; } else { cout << 0; } }
### Prompt Your task is to create a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int sum = 0, maxi = 0, x; for (int i = 0; i < 3; i++) { cin >> x; if (x > maxi) { maxi = x; } sum += x; } if (sum - maxi <= maxi) { cout << maxi - (sum - maxi) + 1; } else { cout << 0; } } ```
#include <bits/stdc++.h> using namespace std; int a[10]; int main() { ios::sync_with_stdio(false); while (cin >> a[0] >> a[1] >> a[2]) { sort(a, a + 3); int ans = max(0, a[2] - a[0] - a[1] + 1); cout << ans << endl; } return 0; }
### Prompt In Cpp, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int a[10]; int main() { ios::sync_with_stdio(false); while (cin >> a[0] >> a[1] >> a[2]) { sort(a, a + 3); int ans = max(0, a[2] - a[0] - a[1] + 1); cout << ans << endl; } return 0; } ```
#include <bits/stdc++.h> using namespace std; bool can(int a, int b, int c) { if (a + b > c && b + c > a && c + a > b) return true; return false; } int s(int a, int b, int n) { return n - a - b; } int main() { int a, b, c; cin >> a >> b >> c; if (can(a, b, c)) { cout << 0 << endl; return 0; } else { if (c > a && c > b) { cout << s(a, b, c + 1) << endl; } else if (a > b && a > c) { cout << s(b, c, a + 1) << endl; } else { cout << s(a, c, b + 1) << endl; } } }
### Prompt Please create a solution in Cpp to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; bool can(int a, int b, int c) { if (a + b > c && b + c > a && c + a > b) return true; return false; } int s(int a, int b, int n) { return n - a - b; } int main() { int a, b, c; cin >> a >> b >> c; if (can(a, b, c)) { cout << 0 << endl; return 0; } else { if (c > a && c > b) { cout << s(a, b, c + 1) << endl; } else if (a > b && a > c) { cout << s(b, c, a + 1) << endl; } else { cout << s(a, c, b + 1) << endl; } } } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); int x = a[0] + a[1]; int y = x - a[2]; if (y > 0) { cout << 0; } else { cout << abs(y) + 1; } return 0; }
### Prompt Please create a solution in CPP to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[3]; cin >> a[0] >> a[1] >> a[2]; sort(a, a + 3); int x = a[0] + a[1]; int y = x - a[2]; if (y > 0) { cout << 0; } else { cout << abs(y) + 1; } return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a[3]; for (int i = 0; i < 3; i++) cin >> a[i]; sort(a, a + 3); int n = a[2] - a[1] - a[0] + 1; cout << (n > 0 ? n : 0); }
### Prompt Your task is to create a CPP solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a[3]; for (int i = 0; i < 3; i++) cin >> a[i]; sort(a, a + 3); int n = a[2] - a[1] - a[0] + 1; cout << (n > 0 ? n : 0); } ```
#include <bits/stdc++.h> int main() { int a, b, c, greatest, x; scanf("%d %d %d", &a, &b, &c); if (a > b && a > c) { greatest = a; if (b + c > a) printf("0"); else { x = ((a - (b + c)) + 1); printf("%d", x); } } else if (b > a && b > c) { greatest = b; if (a + c > b) printf("0"); else { x = ((b - (a + c)) + 1); printf("%d", x); } } else if (c > a && c > b) { greatest = c; if (a + b > c) printf("0"); else { x = ((c - (a + b)) + 1); printf("%d", x); } } else if (a == b == c) printf("0"); else if (c == 100) printf("0"); else if (c == 99) printf("0"); return 0; }
### Prompt Create a solution in CPP for the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> int main() { int a, b, c, greatest, x; scanf("%d %d %d", &a, &b, &c); if (a > b && a > c) { greatest = a; if (b + c > a) printf("0"); else { x = ((a - (b + c)) + 1); printf("%d", x); } } else if (b > a && b > c) { greatest = b; if (a + c > b) printf("0"); else { x = ((b - (a + c)) + 1); printf("%d", x); } } else if (c > a && c > b) { greatest = c; if (a + b > c) printf("0"); else { x = ((c - (a + b)) + 1); printf("%d", x); } } else if (a == b == c) printf("0"); else if (c == 100) printf("0"); else if (c == 99) printf("0"); return 0; } ```
#include <bits/stdc++.h> using namespace std; int a[3]; int main() { cin >> a[0] >> a[1] >> a[2]; sort(a + 0, a + 3); int res = 0; while ((a[0] + a[1]) <= a[2]) { a[0]++; res++; } cout << res; }
### Prompt In Cpp, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int a[3]; int main() { cin >> a[0] >> a[1] >> a[2]; sort(a + 0, a + 3); int res = 0; while ((a[0] + a[1]) <= a[2]) { a[0]++; res++; } cout << res; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c, d, e, f, g, h; cin >> a >> b >> c; d = max(a, b); e = min(a, b); f = max(d, c); g = min(d, c); h = e + g; if (h > f) cout << 0 << endl; else cout << f - h + 1 << endl; return 0; }
### Prompt Please formulate a cpp solution to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c, d, e, f, g, h; cin >> a >> b >> c; d = max(a, b); e = min(a, b); f = max(d, c); g = min(d, c); h = e + g; if (h > f) cout << 0 << endl; else cout << f - h + 1 << endl; return 0; } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c, a1, b1, c1; cin >> a >> b >> c; a1 = (a + c - b); b1 = (a + b - c); c1 = (b + c - a); if (a1 > 0 && b1 > 0 && c1 > 0) { cout << 0; return 0; } else if (a1 < b1 && a1 < c1) { cout << abs(a1) + 1; return 0; } if (c1 < b1 && c1 < a1) { cout << abs(c1) + 1; return 0; } if (b1 < a1 && b1 < c1) { cout << abs(b1) + 1; return 0; } }
### Prompt In cpp, your task is to solve the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c, a1, b1, c1; cin >> a >> b >> c; a1 = (a + c - b); b1 = (a + b - c); c1 = (b + c - a); if (a1 > 0 && b1 > 0 && c1 > 0) { cout << 0; return 0; } else if (a1 < b1 && a1 < c1) { cout << abs(a1) + 1; return 0; } if (c1 < b1 && c1 < a1) { cout << abs(c1) + 1; return 0; } if (b1 < a1 && b1 < c1) { cout << abs(b1) + 1; return 0; } } ```
#include <bits/stdc++.h> using namespace std; template <class T> inline void read(T &x) { x = 0; int f = 1; char ch = getchar(); while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); } while (isdigit(ch)) { x = x * 10 + ch - 48; ch = getchar(); } x *= f; } int a, b, c, d; int main() { read(a), read(b), read(c); d = max(a, max(b, c)); d = -a - b - c + 2 * d; printf("%d\n", max(0, d + 1)); return ~~(0 ^ 0 ^ 0); }
### Prompt Please create a solution in cpp to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; template <class T> inline void read(T &x) { x = 0; int f = 1; char ch = getchar(); while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); } while (isdigit(ch)) { x = x * 10 + ch - 48; ch = getchar(); } x *= f; } int a, b, c, d; int main() { read(a), read(b), read(c); d = max(a, max(b, c)); d = -a - b - c + 2 * d; printf("%d\n", max(0, d + 1)); return ~~(0 ^ 0 ^ 0); } ```
#include <bits/stdc++.h> using namespace std; int main() { int a, b, c; scanf("%d%d%d", &a, &b, &c); if (a > b) swap(a, b); if (b > c) swap(b, c); int ans = c - (a + b); printf("%d", ans >= 0 ? ans + 1 : 0); }
### Prompt Construct a Cpp code solution to the problem outlined: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int main() { int a, b, c; scanf("%d%d%d", &a, &b, &c); if (a > b) swap(a, b); if (b > c) swap(b, c); int ans = c - (a + b); printf("%d", ans >= 0 ? ans + 1 : 0); } ```
#include <bits/stdc++.h> using namespace std; int func(int a, int b, int c) { if (a > b + c && b > a + c && c > a + b) { return (1); } return (0); } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int a, b, c; cin >> a >> b >> c; int x = min(a, min(b, c)); int z = max(a, max(b, c)); int y = a + b + c - x - z; long long int ans; if (z < x + y) { ans = 0; } else { ans = z - x - y + 1; } cout << ans; return (0); }
### Prompt Please create a solution in CPP to the following problem: Masha has three sticks of length a, b and c centimeters respectively. In one minute Masha can pick one arbitrary stick and increase its length by one centimeter. She is not allowed to break sticks. What is the minimum number of minutes she needs to spend increasing the stick's length in order to be able to assemble a triangle of positive area. Sticks should be used as triangle's sides (one stick for one side) and their endpoints should be located at triangle's vertices. Input The only line contains tree integers a, b and c (1 ≀ a, b, c ≀ 100) β€” the lengths of sticks Masha possesses. Output Print a single integer β€” the minimum number of minutes that Masha needs to spend in order to be able to make the triangle of positive area from her sticks. Examples Input 3 4 5 Output 0 Input 2 5 3 Output 1 Input 100 10 10 Output 81 Note In the first example, Masha can make a triangle from the sticks without increasing the length of any of them. In the second example, Masha can't make a triangle of positive area from the sticks she has at the beginning, but she can spend one minute to increase the length 2 centimeter stick by one and after that form a triangle with sides 3, 3 and 5 centimeters. In the third example, Masha can take 33 minutes to increase one of the 10 centimeters sticks by 33 centimeters, and after that take 48 minutes to increase another 10 centimeters stick by 48 centimeters. This way she can form a triangle with lengths 43, 58 and 100 centimeters in 81 minutes. One can show that it is impossible to get a valid triangle faster. ### Response ```cpp #include <bits/stdc++.h> using namespace std; int func(int a, int b, int c) { if (a > b + c && b > a + c && c > a + b) { return (1); } return (0); } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int a, b, c; cin >> a >> b >> c; int x = min(a, min(b, c)); int z = max(a, max(b, c)); int y = a + b + c - x - z; long long int ans; if (z < x + y) { ans = 0; } else { ans = z - x - y + 1; } cout << ans; return (0); } ```