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stringlengths 2
88
| description
stringlengths 31
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| public_tests
dict | private_tests
dict | solution_type
stringclasses 2
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stringclasses 5
values | solution
stringlengths 1
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|
|---|---|---|---|---|---|---|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | #include <bits/stdc++.h>
int comp(const void *a, const void *b) { return *(int *)a - *(int *)b; }
long a[5];
int main(void) {
int n, i, j, k;
scanf("%d\n", &n);
for (i = 0; i < n; i++) {
char c = getchar();
switch (c) {
case 'M':
a[0]++;
break;
case 'A':
a[1]++;
break;
case 'R':
a[2]++;
break;
case 'C':
a[3]++;
break;
case 'H':
a[4]++;
break;
}
while ((c = getchar()) != 10)
;
}
long ans = 0;
for (i = 0; i < n; i++) {
for (j = 0; j < i; j++) {
for (k = 0; k < j; k++) {
ans += a[i] * a[j] * a[k];
}
}
}
printf("%ld\n", ans);
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | N=io.read("n")
local t={M=0,A=0,R=0,C=0,H=0}
for i=1,N do
local s=io.read()
local f=string.sub(s,1,1)
t[f]=t[f]+1 or nil
end
a1 =t[M]*t[A]*t[R]
a2 =t[M]*t[A]*t[C]
a3 =t[M]*t[A]*t[H]
a4 =t[M]*t[R]*t[C]
a5 =t[M]*t[R]*t[H]
a6 =t[M]*t[C]*t[H]
a7 =t[A]*t[R]*t[C]
a8 =t[A]*t[R]*t[H]
a9 =t[A]*t[C]*t[H]
a10=t[R]*t[C]*t[H]
print(a1+a2+a3+a4+a5+a6+a7+a8+a9+a10) |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | from collections import Counter
n=int(input())
name=[]
initial=[]
for _ in range(n):
s=input()
if s[0] in ["M","A","R","C","H"] and not s in name:
name.append(s)
initial.append(s[0])
c=Counter(initial)
num=[v for k,v in c.items()]
if len(num)<3:
print(0)
else:
ans=0
for i in range(len(num)-2):
for j in range(i+1,len(num)-1):
for k in range(j+1,len(num)):
#print(i,j,k)
ans+=num[i]*num[j]*num[k]
print(ans)
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
string s[100000];
int cnt[5] = {};
char c[5] = {'M', 'A', 'R', 'C', 'H'};
long long int ans = 0;
cin >> n;
for (int i = 0; i < n; i++) cin >> s[i];
for (int i = 0; i < n; i++) {
for (int j = 0; j < 5; j++) {
if (s[i][0] == c[j]) {
cnt[j]++;
break;
}
}
}
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += cnt[i] * cnt[j] * cnt[k];
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace AtCoder
{
static class Program
{
static void Main()
{
var N = GetInt();
Dictionary<char, int> march = new Dictionary<char, int>();
var C = new char[] { 'M', 'A', 'R', 'C', 'H' };
foreach (var c in C)
{
march.Add(c, 0);
}
var S = new char[N];
for(int i = 0; i < N; i++)
{
S[i] = GetStr()[0];
}
foreach(var s in S)
{
if (march.ContainsKey(s)) march[s]++;
}
long ans = 0;
var keys = march.Keys.ToArray();
for(int i = 0; i < 3; i++)
{
for(int j = i + 1; j < 4; j++)
{
for (int k = j + 1; k < 5; k++)
{
ans += march[keys[i]] * march[keys[j]] * march[keys[k]];
}
}
}
Console.WriteLine(ans);
Console.ReadKey();
}
static void Solve()
{
}
static public string GetStr() { return Console.ReadLine().Trim(); }
static public int GetInt() { return int.Parse(Console.ReadLine()); }
static public long GetLong() { return long.Parse(Console.ReadLine()); }
static public string[] GetStrArray() { return Console.ReadLine().Split(' '); }
static public int[] GetIntArray() { return Console.ReadLine().Split(' ').Select(int.Parse).ToArray(); }
static public long[] GetLongArray() { return Console.ReadLine().Split(' ').Select(long.Parse).ToArray(); }
static public char[] GetCharArray() { return Console.ReadLine().Split(' ').Select(char.Parse).ToArray(); }
static public List<double> GetDoubleList() { return Console.ReadLine().Split(' ').Select(double.Parse).ToList(); }
static public void WriteObjects<T>(IEnumerable<T> values) { foreach (var o in values) { Console.Write(o + " "); } }
static string yesno(this bool b) { return b ? "yes" : "no"; }
static string YesNo(this bool b) { return b ? "Yes" : "No"; }
static string YESNO(this bool b) { return b ? "YES" : "NO"; }
static bool eq<T, U>() => typeof(T).Equals(typeof(U));
static T ct<T, U>(U a) => (T)Convert.ChangeType(a, typeof(T));
static T cv<T>(string s) => eq<T, int>() ? ct<T, int>(int.Parse(s))
: eq<T, long>() ? ct<T, long>(long.Parse(s))
: eq<T, double>() ? ct<T, double>(double.Parse(s))
: eq<T, char>() ? ct<T, char>(s[0])
: ct<T, string>(s);
static void Multi<T>(out T a) => a = cv<T>(GetStr());
static void Multi<T, U>(out T a, out U b)
{
var ar = GetStrArray(); a = cv<T>(ar[0]); b = cv<U>(ar[1]);
}
static void Multi<T, U, V>(out T a, out U b, out V c)
{
var ar = GetStrArray(); a = cv<T>(ar[0]); b = cv<U>(ar[1]); c = cv<V>(ar[2]);
}
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | from itertools import combinations
N = int(input())
S = Counter()
for _ in range(N):
S[input()[0]] += 1
print(sum([S[p0]*S[p1]*S[p2] for p0,p1,p2 in combinations('MARCH', 3)])) |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | N=int(input())
S=[input() for _ in range(N)]
S.sort()
S.append('0')
dp=[0]*5
for i in range(N+1):
if i==N:
continue
if S[i]==S[i+1]:
continue
if S[i][0]=='M':
dp[0]+=1
elif S[i][0]=='A':
dp[1]+=1
elif S[i][0]=='R':
dp[2]+=1
elif S[i][0]=='C':
dp[3]+=1
elif S[i][0]=='H':
dp[4]+=1
ans=0
for i in range(2):
for j in range(i+1,4):
for k in range(j+1,5):
ans+=dp[i]*dp[j]*dp[k]
print(ans) |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
int main() {
ios_base::sync_with_stdio(false);
int n;
cin >> n;
int a[7];
for (int i = 0; i <= 6; ++i) a[i] = 0;
for (int i = 1; i <= n; ++i) {
string s;
cin >> s;
if (s[0] == 'M') a[1]++;
if (s[0] == 'A') a[2]++;
if (s[0] == 'R') a[3]++;
if (s[0] == 'C') a[4]++;
if (s[0] == 'H') a[5]++;
}
vector<pair<int, int> > v;
if (a[1]) v.push_back({a[1], 1});
if (a[2]) v.push_back({a[2], 2});
if (a[3]) v.push_back({a[3], 3});
if (a[4]) v.push_back({a[4], 4});
if (a[5]) v.push_back({a[5], 5});
long long ans = 1;
if (v.size() < 3) {
cout << 0 << '\n';
return 0;
}
do {
ans += (1LL * v[0].first * v[1].first * v[2].first);
} while (next_permutation((v).begin(), (v).end()));
ans /= 6LL;
cout << ans << '\n';
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
int N;
scanf("%d", &N);
char c, dummy;
int cnt[5] = {0};
for (int i = 0; i < N; i++) {
scanf(" %c%*[^\n]", &c);
switch (c) {
case 'M':
cnt[0]++;
break;
case 'A':
cnt[1]++;
break;
case 'R':
cnt[2]++;
break;
case 'C':
cnt[3]++;
break;
case 'H':
cnt[4]++;
break;
default:
break;
}
}
long long result = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
result += cnt[i] * cnt[j] * cnt[k];
}
}
}
printf("%lld", result);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<string> s(n);
for (int i = 0; i < n; i++) cin >> s[i];
vector<pair<char, int> > p(5);
p[0] = make_pair('M', 0);
p[1] = make_pair('A', 0);
p[2] = make_pair('R', 0);
p[3] = make_pair('C', 0);
p[4] = make_pair('H', 0);
int use = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < 5; j++) {
if (s[i].at(0) == p[j].first) {
if (p[j].second == 0) use++;
p[j].second++;
}
}
}
int ans = 0;
if (use >= 3) {
for (int i = 0; i < 5; i++) {
if (p[i].second > 1) ans += (p[i].second - 1) * 3;
}
if (use == 4) ans += 4;
if (use == 5) ans += 10;
} else {
cout << 0 << endl;
return 0;
}
std::cout << ans << '\n';
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<map>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep2(i,j,n) for(int i=(j)+1;i<(n);i++)
int m = 0;
int y;
int k1(string a) {
if (a == "M") {
m++;
return 1;
}
else if (a == "A") {
m++;
return 2;
}
else if (a == "R") {
m++;
return 3;
}
else if (a == "C") {
m++;
return 4;
}
else if (a == "H") {
m++;
return 5;
}
else {
return 0;
}
}
int main() {
long int n;
cin >> n;
vector<string> s(n);
rep(i, n) {
cin >> s[i];
}
vector<int> x(n);
rep(i, n) {
string a = s[i];
a = a[0];
x[i] = k1(a);
}
if (m < 3) {
cout << 0 << endl;
return 0;
}
sort(x.begin(), x.end());
int ans = 0;
rep(i, n) {
if (x[i] == 0) {
continue;
}
rep2(j, i, n) {
if (x[j] == 0 || x[j] == x[i]) {
continue;
}
rep(k, j, n) {
if (x[k] == 0) {
continue;
}
else if (x[k] == x[i] || x[k] == x[j]) {
continue;
}
ans++;
}
}
}
cout << ans / 2 << endl;
return 0;
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int N = scn.nextInt();
int[] num = new int[5];
scn.nextLine();
for (int i = 0; i < N; i++) {
char[] buf = scn.nextLine().toCharArray();
switch (buf[0]) {
case 'M':
num[0]++;
break;
case 'A':
num[1]++;
break;
case 'R':
num[2]++;
break;
case 'C':
num[3]++;
break;
case 'H':
num[4]++;
break;
default:
break;
}
}
long ans = 0;
for(int i = 0;i < 3;i++) {
for(int j = i+1;j < 4;j++) {
for(int k = j+1;k < 5;k++) {
ans += num[i] * num[j] * num[k];
}
}
}
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | math
N = int(input())
S = [input() for i in range(N)]
M=0
A=0
R=0
C=0
H=0
cont1=0
cont2=0
cont3=0
cont4=0
cont5=0
count=0
for i in range(N):
if S[i] == 'M':
M+=1
elif S[i] == 'A':
A+=1
elif S[i] == 'C':
C+=1
elif S[i] == 'R':
R+=1
elif S[i] == 'H':
H+=1
#print(M,A,R,C,H)
num = M+A+R+C+H
if num>3:
num1 = math.factorial(num)
num11=math.factorial(num-3)*3*2
num2= num1//num11
if M>1:
cont1=M
elif A>1:
cont2=A
elif R>1:
cont3=R
elif C>1:
cont4=C
elif H>1:
cont5=H
cont =cont1+cont2+cont3+cont4+cont5
count=num-cont
print(num2-count)
# print(num2,num,cont)
else:
print(0)
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace AtCoder
{
class Program
{
static void Main(string[] args)
{
int n = int.Parse(Console.ReadLine());
var dic = new Dictionary<char, int>();
string s = string.Empty;
var march = new char[] { 'M', 'A', 'R', 'C', 'H' };
foreach(char c in march)
{
dic[c] = 0;
}
for(int i = 0; i < n; i++)
{
s = Console.ReadLine();
if (!march.Contains(s.First())) continue;
dic[s.First()]++;
}
var check = new int[] { 7, 11, 13, 14, 19, 21, 22, 25, 26, 28 };
int comb = 1;
int sum = 0;
for(int i = 7; i <= 28 ; i++)
{
if (!check.Contains(i)) continue;
comb = 1;
for(int j = 0; j < 5; j++)
{
if ((i >> j & 1) == 1)
{
comb *= dic[march[j]];
}
}
sum += comb;
}
Console.WriteLine(sum);
}
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include"iostream"
2.#include"string"
3.using namespace std;
4.
5.int main()
6.{
7. unsigned long long n,k,m,a,r,c,h;
8. m=a=r=c=h= 0;
9. cin >> n;
10.
11. do {
12. string x,ch;
13. cin >> x;
14. ch = x[0];
15. if (ch == "M")m = m + 1;
16. else if (ch == "A") a = a + 1;
17. else if (ch == "R")r = r + 1;
18. else if (ch == "C")c = c + 1;
19. else if (ch == "H")h = h + 1;
20.
21. n=n-1;
22.
23. }while (n >= 1);
24.
25. unsigned long long l = m * a*r + m * a*c + m * a*h + m * r*c + m * r*h + m * c*h + a * r*c + a * r*h + a * c*h + r * c*h;
26. cout <<l;
27.}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
inline long long read() {
long long tmp = 0;
char c = getchar(), f = 1;
for (; c < '0' || '9' < c; c = getchar())
if (c == '-') f = -1;
for (; '0' <= c && c <= '9'; c = getchar())
tmp = (tmp << 3) + (tmp << 1) + c - '0';
return tmp * f;
}
using namespace std;
int cnt[256];
char s[20];
int main() {
int i, j, k, n = read();
for (i = 1; i <= n; i++) {
scanf("%s", s);
if (s[0] == 'A' || s[0] == 'R' || s[0] == 'M' || s[0] == 'C' || s[0] == 'H')
++cnt[(int)s[0]];
}
long long ans = 0;
for (i = 0; i < 256; i++)
for (j = i + 1; j < 256; j++)
for (k = j + 1; k < 256; k++) ans += cnt[i] * cnt[j] * cnt[k];
printf("%lld\n", ans);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> march(5);
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] == 'M') march[0]++;
if (s[0] == 'A') march[1]++;
if (s[0] == 'R') march[2]++;
if (s[0] == 'C') march[3]++;
if (s[0] == 'H') march[4]++;
}
int ans = 0;
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int s = j + 1; s < 5; s++) {
ans += march[i] * march[j] * march[s];
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;
using ll = long long;
int MARCH(vector<string> vec, int cnt) {
int M = 0, A = 0, R = 0, C = 0, H = 0;
for (int i = 0; i < (int)vec.size(); i++) {
if (vec.at(i).at(0) == 'M') M++;
if (vec.at(i).at(0) == 'A') A++;
if (vec.at(i).at(0) == 'R') R++;
if (vec.at(i).at(0) == 'C') C++;
if (vec.at(i).at(0) == 'H') H++;
}
if (M > 0) cnt++;
if (A > 0) cnt++;
if (R > 0) cnt++;
if (C > 0) cnt++;
if (H > 0) cnt++;
return cnt;
}
int combination(int n, int k) {
double ans = 1;
if (k == 1) {
return n / k;
}
ans = (double)n / k * combination(n - 1, k - 1);
return ans;
}
int main() {
int n;
cin >> n;
string s = "MARCH";
vi c(5);
for (int i = 0; i < n; i++) {
string t;
cin >> t;
for (int j = 0; j < 5; j++) {
if (t.at(0) == s.at(j)) {
c.at(j)++;
}
}
}
int ans = 0;
for (int i = 0; i < 5; i++)
for (int j = i + 1; j < 5; j++)
for (int k = j + 1; k < 5; k++) {
ans += c.at(i) * c.at(j) * c.at(k);
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] nums = {0, 0, 0, 0, 0};
for(int i = 0; i < n; i++) {
String str = sc.next();
switch(str.charAt(0)) {
case 'M': nums[0]++; break;
case 'A': nums[1]++; break;
case 'R': nums[2]++; break;
case 'C': nums[3]++; break;
case 'H': nums[4]++; break;
}
}
int type = 0;
for(int i = 0; i < 5; i++) {
if(nums[i] != 0)
type++;
}
if(type < 3)
System.out.println(0);
else if(type == 3) {
long cnt = 1;
for(int i = 0; i < 5; i++) {
if(nums[i] != 0)
cnt*=nums[i];
}
System.out.println(cnt);
} else {
long cnt = 0;
for(int i = 0; i < 5; i++) {
if(nums[i] == 0) continue;
long buf = 1;
for(int j = 0; j < 5; j++) {
if(i == j) continue;
if(nums[j] != 0)
buf *= nums[j];
}
cnt += buf;
}
System.out.println(cnt);
}
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main(void) {
int n;
cin >> n;
string name;
int m, a, r, c, h = 0;
for (int i = 0; i < n; i++) {
cin >> name;
if (name[0] = 'M') m++;
if (name[0] = 'A') a++;
if (name[0] = 'R') r++;
if (name[0] = 'C') c++;
if (name[0] = 'H') h++;
}
cout << m * a * r + m * a * c + m * a * h + m * r * c + m * r * h +
m * c * h + a * r * c + a * r * h + a * c * h + r * c * h
<< endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
int m, a, r, c, h;
m = a = r = c = h = 0;
string tmp;
for (int i = 0; i < (int)(N); i++) {
cin >> tmp;
if (tmp.at(0) == 'M')
m++;
else if (tmp.at(0) == 'A')
a++;
else if (tmp.at(0) == 'R')
r++;
else if (tmp.at(0) == 'C')
c++;
else if (tmp.at(0) == 'H')
h++;
}
long ans = m * a * (r + c + h) + m * r * (c + h) + m * c * h +
a * r * (c + h) + a * c * h + r * c * h;
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main(){
long long march[5]={},ans=1;
bool exist=false;
string s;
cin>>n;
for(int i=0;i<n;i++){
cin>>s;
switch(s[0]){
case 'M':
march[0]++;
break;
case 'A':
march[1]++;
break;
case 'R':
march[2]++;
break;
case 'C':
march[3]++;
break;
case 'H':
march[4]++;
break;
default:
break;
}
}
for(int i=0;i<5;i++){
if(march[i]){
exist=true;
ans*=march[i];
}
}
cout<<(exist?ans:0)<<endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m = 0, a = 0, r = 0, c = 0, h = 0;
cin >> n;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] == "M") {
m++;
} else if (s[0] == "A") {
a++;
} else if (s[0] == "R") {
r++;
} else if (s[0] == "C") {
c++;
} else if (s[0] == "H") {
h++;
}
}
cout << m*a*r + m*a*c + m*a*h + m*r*c + m*r*h + m*c*h + a*r*c + a*r*h + a*c*h + r*c*h << endl;
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
int main() {
ll n;
cin >> n;
vector<ll> v(n, 0);
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s.at(0) == 'M') {
v.at(0)++;
} else if (s.at(0) == 'A') {
v.at(1)++;
} else if (s.at(0) == 'R') {
v.at(2)++;
} else if (s.at(0) == 'C') {
v.at(3)++;
} else if (s.at(0) == 'H') {
v.at(4)++;
}
}
ll ans = 0;
for (int bit = 0; bit < (1 << 5); bit++) {
ll sum = 1;
int a = 0;
for (int i = 0; i < 5; i++) {
if (bit & (1 << i)) {
a++;
sum *= v.at(i);
}
}
if (a == 3) {
ans += sum;
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
string s;
long long MARCH[5] = {};
for (long long i = 0; i < n; i++) {
s.shrink_to_fit();
cin >> s;
switch (s[0]) {
case 'M':
MARCH[0]++;
break;
case 'A':
MARCH[1]++;
break;
case 'R':
MARCH[2]++;
break;
case 'C':
MARCH[3]++;
break;
case 'H':
MARCH[4]++;
break;
}
}
long long ans = 0;
for (long long i = 0; i < n; i++) {
for (long long j = i + 1; j < n; j++) {
for (long long k = j + 1; k < n; k++) {
ans += MARCH[i] * MARCH[j] * MARCH[k];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
int N;
cin >> N;
vector<string> S(N);
for (int i = 0; i < (N); i++) {
cin >> S[i];
}
int march[5] = {};
for (int i = 0; i < (N); i++) {
if (S[i][0] == 'M') {
march[0]++;
} else if (S[i][0] == 'A') {
march[1]++;
} else if (S[i][0] == 'R') {
march[2]++;
} else if (S[i][0] == 'C') {
march[3]++;
} else if (S[i][0] == 'H') {
march[4]++;
}
}
long long ans = 0;
for (int i = 0; i < (5); i++) {
for (int j = 0; j < (5); j++) {
for (int k = 0; k < (5); k++) {
if (i != j && j != k && i != k && i < j && j < k) {
ans += march[i] * march[j] * march[k];
}
}
}
}
cout << ans << endl;
;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int factorial(int x) {
int tmp = x;
for (int i = x - 1; i > 0; i--) tmp *= i;
return tmp;
}
int main() {
unsigned long int N;
cin >> N;
char name[N][15];
char c[5] = {'M', 'A', 'R', 'C', 'H'};
for (int i = 0; i < N; i++) cin >> name[i];
int count = 0;
for (int i = 0; i < N; i++) {
for (int j = i + 1; j < N; j++) {
for (int k = j + 1; k < N; k++) {
int flag[3] = {0, 0, 0};
if (name[i][0] != name[j][0] && name[i][0] != name[k][0] &&
name[k][0] != name[j][0]) {
for (int m = 0; m < 5; m++) {
if (flag[0] == 0 && name[i][0] == c[m]) flag[0] = 1;
if (flag[1] == 0 && name[j][0] == c[m]) flag[1] = 1;
if (flag[2] == 0 && name[k][0] == c[m]) flag[2] = 1;
}
if (flag[0] + flag[1] + flag[2] == 3) count++;
}
}
}
}
cout << count << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long n, ans;
map<char, int> mp;
char s[1000];
int a[10];
int main() {
cin >> n;
mp['M'] = 1;
mp['A'] = 2;
mp['R'] = 3;
mp['C'] = 4;
mp['H'] = 5;
for (int i = 1; i <= n; ++i) {
scanf("%s", s + 1);
if (!mp[s[1]]) continue;
a[mp[s[1]]]++;
}
for (int i = 1; i <= 5; ++i) {
for (int j = i + 1; j <= 5; ++j) {
if (i == j) continue;
for (int k = j + 1; k <= 5; ++k) {
if (k == i || k == j) continue;
ans += a[k] * a[j] * a[i];
}
}
}
cout << ans;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, count[5];
long long ans = 0;
string s;
for (int i = 0; i < 5; i++) count[i] = 0;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> s;
if (s[0] == 'M') count[0]++;
if (s[0] == 'A') count[1]++;
if (s[0] == 'R') count[2]++;
if (s[0] == 'C') count[3]++;
if (s[0] == 'H') count[4]++;
}
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
ans += count[i] * count[j] * count[k];
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
inline int read() {
char c;
int s = 0, t = 1;
while (!isdigit(c = getchar()))
if (c == '-') t = -1;
do {
s = s * 10 + c - '0';
} while (isdigit(c = getchar()));
return s * t;
}
inline long long readl() {
char c;
long long s = 0;
int t = 1;
while (!isdigit(c = getchar()))
if (c == '-') t = -1;
do {
s = s * 10 + c - '0';
} while (isdigit(c = getchar()));
return s * t;
}
int a[100];
int main() {
int n;
n = read();
for (int i = 1; i <= n; i++) {
char s[20];
scanf("%s", s);
if (s[0] == 'M') a[1]++;
if (s[0] == 'A') a[2]++;
if (s[0] == 'R') a[3]++;
if (s[0] == 'C') a[4]++;
if (s[0] == 'H') a[5]++;
}
long long ans = 0;
for (int i = 1; i <= 5; i++)
for (int j = i + 1; j <= 5; j++)
for (int k = j + 1; k <= 5; k++) ans += a[i] * a[j] * a[k] * 1ll;
printf("%lld\n", ans);
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, M = 0, A = 0, R = 0, C = 0, H = 0, i, a[10];
long long s = 0LL;
string L;
cin >> n;
for (i = 0; i < n; i++) {
cin >> L;
switch (L[0]) {
case 'M':
M++;
break;
case 'A':
A++;
break;
case 'R':
R++;
break;
case 'C':
C++;
break;
case 'H':
H++;
break;
default:
break;
}
}
a[0] = M * A * R;
a[1] = M * A * C;
a[2] = M * A * H;
a[3] = M * R * C;
a[4] = M * R * H;
a[5] = M * C * H;
a[6] = A * R * C;
a[7] = A * R * H;
a[8] = A * C * H;
a[9] = R * C * H;
for (i = 0; i < 10; i++) {
s += a[i];
}
cout << s << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | #include <bits/stdc++.h>
int main(void) {
int N, i, j, k;
int count[5] = {0};
long long int sum = 0;
char S[12];
scanf("%d", &N);
for (i = 0; i < N; i++) {
scanf("%s", S);
switch (S[0]) {
case 'M':
count[0]++;
break;
case 'A':
count[1]++;
break;
case 'R':
count[2]++;
break;
case 'C':
count[3]++;
break;
case 'H':
count[4]++;
break;
}
}
for (i = 0; i < 3; i++) {
for (j = i + 1; j < 4; j++) {
for (k = j + 1; k < 5; k++) {
sum += count[i] * count[j] * count[k];
}
}
}
printf("%lld\n", sum);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java |
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] list = new int[5];
for(int i = 0; i < 5; i++) {
list[i] = 0;
}
for(int i = 0; i < n; i++) {
String s = sc.next();
if(s.charAt(0) == 'M')
list[0]++;
else if(s.charAt(0) == 'A')
list[1]++;
else if(s.charAt(0) == 'R')
list[2]++;
else if(s.charAt(0) == 'C')
list[3]++;
else if(s.charAt(0) == 'H')
list[4]++;
}
long ans = 0;
for(int i = 0; i < 5; i++) {
for(int j = i + 1; j < 5; j++) {
for(int k = j + 1; k < 5; k++) {
ans += list[i] * list[j] * list[k];
}
}
}
System.out.println(ans);
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include<iostream>
using namespace std;
int main(){
int N;
cin >> N;
char name[100000][11];
for(int i=0;i<N;i++){
cin >> a[i];
}
int m=0,a=0,r=0,c=0,h=0;
for(int i=0;i<N;i++){
switch(a[i][1]){
case 'M':
m++;
break;
case 'A':
a++;
break;
case 'R':
r++;
break;
case 'C':
c++;
break;
case 'H':
h++;
break;
default:
break;
}
}
long long ans;
ans=m*a*r*c*h+0;
cout << ans;
return 0;
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <class T>
bool chmax(T &a, const T &b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
if (b < a) {
a = b;
return 1;
}
return 0;
}
int dy[] = {0, 0, 1, -1};
int dx[] = {1, -1, 0, 0};
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
int n, m, a, r, c, h;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
cin >> n;
for (int i = 0; i < (n); ++i) {
string s;
cin >> s;
char ch = s[0];
if (ch == 'M')
m++;
else if (ch == 'A')
a++;
else if (ch == 'R')
r++;
else if (ch == 'C')
c++;
else if (ch == 'H')
h++;
}
int ans = m * a * r + m * a * c + m * a * h + m * r * c + m * r * h +
m * c * h + a * r * c + a * r * h + a * c * h + r * c * h;
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define int long long
int q[200];
main(){
int n,i,c,k,l=0;
cin>>n;
string s;
for (i=0;i<n;++i){
cin>>s;
q[s[0]]++;
}
int r=1;
c=q['M']+q['A']+q['R']+q['C']+q['H'];
if (q['M']>1){
l+=q['A']+q['R']+q['C']+q['H'];
if (l==0){
cout <<0;
return 0;
}
}
if (q['A']>1){
l+=q['M']+q['R']+q['C']+q['H'];
if (l==0){
cout <<0;
return 0;
}
}
if (q['R']>1){
l+=q['M']+q['A']+q['C']+q['H'];
if (l==0){
cout <<0;
return 0;
}
}
if (q['C']>1){
l+=q['M']+q['A']+q['R']+q['H'];
if (l==0){
cout <<0;
return 0;
}
}
if (q['H']>1){
l+=q['M']+q['A']+q['R']+q['C'];
if (l==0){
cout <<0;
return 0;
}
}
if (c-3>3){
k=1;
int pr=0;
for (i=c-2;i<=c;++i){
k*=i;
if (k%6==0){
k/=6;
pr=1;
}
}
if (pr==0)
r=6;
}
else {
k=1;
for (i=4;i<=c;++i){
k*=i;
}
for (i=2;i<=c-3;++i){
r*=i;
}
}
if (c-l<3)
cout <<0;
else
cout <<l;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MX = 100005, INF = 1001001001;
const long long int LINF = 1e18;
const double eps = 1e-10;
const int DIV = 1e9 + 7;
int n;
vector<string> s;
int main() {
cin >> n;
s.resize(n);
for (int i = 0; i < (n); ++i) cin >> s[i];
map<char, set<string>> namesMap;
for (int i = 0; i < (n); ++i) {
namesMap[s[i][0]].insert(s[i]);
}
vector<int> candsNum;
candsNum.push_back(namesMap['M'].size());
candsNum.push_back(namesMap['A'].size());
candsNum.push_back(namesMap['R'].size());
candsNum.push_back(namesMap['C'].size());
candsNum.push_back(namesMap['H'].size());
long long int ans = 0;
for (int v1 = 0; v1 < 3; v1++) {
for (int v2 = v1 + 1; v2 < 4; v2++) {
for (int v3 = v2 + 1; v3 < 5; v3++) {
ans += (candsNum[v1] * candsNum[v2] * candsNum[v3]);
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | #include <bits/stdc++.h>
int main() {
long long ans = 1;
int march[5] = {0};
int n, sum = 0, el = 0;
scanf("%d", &n);
char s[15];
for (int i = 0; i < n; i++) {
scanf("%s", s);
switch (s[0]) {
case 'M':
march[0]++;
break;
case 'A':
march[1]++;
break;
case 'R':
march[2]++;
break;
case 'C':
march[3]++;
break;
case 'H':
march[4]++;
break;
default:
break;
}
}
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
ans += march[i] * march[j] * march[k];
}
}
}
printf("%lld", ans);
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | a
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long int pow(int x, int n) { return n == 0 ? 1 : x * pow(x, n - 1); }
int n;
int nums[5];
char str[] = "MARCH";
long long int dp[6][4];
int main() {
cin >> n;
for (int i = 0; i < (n); i++) {
string s;
cin >> s;
for (int j = 0; j < (5); j++) {
if (s[0] == str[j]) nums[j]++;
}
}
for (int i = 0; i < (5); i++)
if (1) fprintf(stderr, "%d ", nums[i]);
cerr << endl;
long long int ans = 0;
for (int i = 0; i < (5); i++) {
for (int j = (i + 1); j < (5); j++) {
for (int k = (j + 1); k < (5); k++) {
ans += nums[i] * nums[j] * nums[k];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, ans = 0;
string temp;
char target[5] = {'M', 'A', 'R', 'C', 'H'};
int count[5] = {0, 0, 0, 0, 0};
cin >> n;
for (int i = 0; i < n; i++) {
cin >> temp;
for (int j = 0; j < 5; j++) {
if (temp[0] == target[j]) count[j]++;
}
}
ans = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += count[i] * count[j] * count[k];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
int main() {
int N;
std::cin >> N;
std::array<int, 5> s{};
for (auto i = 0; i < N; ++i) {
std::string name;
std::cin >> name;
switch (name[0]) {
case 'M':
s[0]++;
break;
case 'A':
s[1]++;
break;
case 'R':
s[2]++;
break;
case 'C':
s[3]++;
break;
case 'H':
s[4]++;
break;
default:
break;
}
}
std::cout << s[0] * s[1] * s[2] + s[0] * s[1] * s[3] + s[0] * s[1] * s[4] +
s[0] * s[2] * s[3] + s[0] * s[2] * s[4] +
s[0] * s[3] * s[4] + s[1] * s[2] * s[3] +
s[1] * s[2] * s[4] + s[1] * s[3] * s[4] + s[2] * s[3] * s[4]
<< std::endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
char S[11];
scanf("%d", &N);
int M, A, R, C, H;
M = A = R = C = H = 0;
for (int i = 1; i <= N; ++i) {
scanf("%s", S);
if (S[0] == 'M')
M++;
else if (S[0] == 'A')
A++;
else if (S[0] == 'R')
R++;
else if (S[0] == 'C')
C++;
else if (S[0] == 'H')
H++;
else {
}
}
long ans = M * A * (R + C + H) + A * R * (C + H) + M * R * (C + H) +
(A + M + R) * C * H;
printf("%ld\n", ans);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
string t = "MARCH";
int cnt[5] = {};
for (int i = 0; i < n; i++) {
string s;
cin >> s;
for (int j = 0; j < 5; j++) {
if (s[0] == t[j]) cnt[j]++;
}
}
long long ans = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += cnt[i] * cnt[j] * cnt[k];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
char name[100000][11];
int m = 0, a = 0, r = 0, c = 0, h = 0;
for (int i = 0; i < N; i++) {
cin >> name[i];
switch (name[i][0]) {
case 'M':
m++;
break;
case 'A':
a++;
break;
case 'R':
r++;
break;
case 'C':
c++;
break;
case 'H':
h++;
break;
default:
break;
}
}
long long ans;
ans = m * (a * r + a * c + a * h + r * c + r * h + c * h) +
a * (r * c + r * h + c * h) + r * c * h + 0;
cout << ans;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | using System;
using System.Collections.Generic;
using System.Linq;
using static System.Math;
using static AtCoderTemplate.MyExtensions;
using static AtCoderTemplate.MyInputOutputs;
using static AtCoderTemplate.MyNumericFunctions;
namespace AtCoderTemplate {
public class Program {
public static void Main (string[] args) {
var N = ReadInt ();
var S = new List<string> ();
foreach (var i in Enumerable.Range (0, N)) {
S.Add (Console.ReadLine ());
}
var M = S.Where (Si => Si[0] == 'M').Count ();
var A = S.Where (Si => Si[0] == 'A').Count ();
var R = S.Where (Si => Si[0] == 'R').Count ();
var C = S.Where (Si => Si[0] == 'C').Count ();
var H = S.Where (Si => Si[0] == 'H').Count ();
var not0 = new List<int> { M, A, R, C, H }.Where (i => i != 0).ToList ();
if (not0.Count < 3) {
Print (0);
} else {
var lastIx = not0.Count - 1;
var ans = Enumerable.Range (0, lastIx - 2 + 1)
.SelectMany (a => Enumerable.Range (a + 1, lastIx - 1 - (a + 1) + 1)
.SelectMany (b => Enumerable.Range (b + 1, lastIx - (b + 1) + 1)
.Select (c => {
// Print ($"{a}, {b}, {c}");
return not0[a] * not0[b] * not0[c];
})))
.Sum ();
Print (ans);
}
}
}
static class MyInputOutputs {
/* Input & Output*/
public static int ReadInt () {
return int.Parse (Console.ReadLine ());
}
public static long ReadLong () {
return long.Parse (Console.ReadLine ());
}
public static List<int> ReadInts () {
return Console.ReadLine ().Split (' ').Select (c => int.Parse (c)).ToList ();
}
public static List<long> ReadLongs () {
return Console.ReadLine ().Split (' ').Select (c => long.Parse (c)).ToList ();
}
public static List<List<int>> ReadIntColumns (int n) {
/*
入力例
A1 B1
A2 B2
...
An Bn
出力例
[[A1,A2,...,An], [B1,B2,...,Bn]]
*/
var rows = Enumerable.Range (0, n).Select (i => ReadInts ()).ToList ();
var m = rows.FirstOrDefault ()?.Count () ?? 0;
return Enumerable.Range (0, m).Select (i => rows.Select (items => items[i]).ToList ()).ToList ();
}
public static List<List<long>> ReadLongColumns (int n) {
/*
入力例
A1 B1
A2 B2
...
An Bn
出力例
[[A1,A2,...,An], [B1,B2,...,Bn]]
*/
var rows = Enumerable.Range (0, n).Select (i => ReadLongs ()).ToList ();
var m = rows.FirstOrDefault ()?.Count () ?? 0;
return Enumerable.Range (0, m).Select (i => rows.Select (items => items[i]).ToList ()).ToList ();
}
public static void Print<T> (T item) {
Console.WriteLine (item);
}
public static void PrintIf<T1, T2> (bool condition, T1 trueResult, T2 falseResult) {
if (condition) {
Console.WriteLine (trueResult);
} else {
Console.WriteLine (falseResult);
}
}
public static void PrintRow<T> (IEnumerable<T> list) {
/* 横ベクトルで表示
A B C D ...
*/
if (!list.IsEmpty ()) {
Console.Write (list.First ());
foreach (var item in list.Skip (1)) {
Console.Write ($" {item}");
}
}
Console.Write ("\n");
}
public static void PrintColomn<T> (IEnumerable<T> list) {
/* 縦ベクトルで表示
A
B
C
D
...
*/
foreach (var item in list) {
Console.WriteLine (item);
}
}
public static void Print2DArray<T> (IEnumerable<IEnumerable<T>> sources) {
foreach (var row in sources) {
PrintRow (row);
}
}
}
static class MyNumericFunctions {
public static bool isEven (int a) {
return a % 2 == 0;
}
public static bool isEven (long a) {
return a % 2 == 0;
}
public static bool isOdd (int a) {
return !isEven (a);
}
public static bool isOdd (long a) {
return !isEven (a);
}
public static long nPk (int n, int k) {
if (k > n) {
return 0;
} else {
return Enumerable.Range (n - k + 1, k).Aggregate ((long) 1, ((i, m) => i * m));
}
}
public static long Fact (int n) {
return nPk (n, n);
}
public static long nCk (int n, int k) {
if (k > n) {
return 0;
} else {
return nPk (n, k) / Fact (k);
}
}
// 最大公約数
public static long GCD (long m, long n) {
if (m < n) {
return GCD (n, m);
} else if (n == 0) {
return m;
} else {
return GCD (n, m % n);
}
}
// 最小公倍数
public static long LCM (long m, long n) {
if (m < n) {
return LCM (n, m);
} else {
return (m * n) / GCD (m, n);
}
}
}
static class MyExtensions {
// AppendとPrependが、.NET Standard 1.6からの追加で、Mono 4.6.2 はそれに対応して仕様はあるが、実装がない
public static IEnumerable<T> Append<T> (this IEnumerable<T> source, T element) {
return source.Concat (Enumerable.Repeat (element, 1));
}
public static IEnumerable<T> Prepend<T> (this IEnumerable<T> source, T element) {
return Enumerable.Repeat (element, 1).Concat (source);
}
// TakeLastとSkipLastが、.Net Standard 2.1からの追加で、Mono 4.6.2 はそれに対応していない
public static IEnumerable<T> TakeLast<T> (this IEnumerable<T> source, int count) {
return source.Skip (source.Count () - count);
}
public static IEnumerable<T> SkipLast<T> (this IEnumerable<T> source, int count) {
return source.Take (source.Count () - count);
}
public static bool IsEmpty<T> (this IEnumerable<T> source) {
return !source.Any ();
}
/// <summary>
/// インデックスiの位置の要素からk個取り除く
/// O(N)
/// </summary>
public static IEnumerable<T> TakeAwayRange<T> (this IEnumerable<T> source, int i, int count) {
return source.Take (i).Concat (source.Skip (i + count));
}
/// <summary>
/// インデックスiの位置の要素を取り除く
/// O(N)
/// </summary>
public static IEnumerable<T> TakeAwayAt<T> (this IEnumerable<T> source, int i) {
return source.TakeAwayRange (i, 1);
}
/// <summary>
/// インデックスiの位置にシーケンスを挿入する
/// O(N + K)
/// </summary>
public static IEnumerable<T> InsertEnumAt<T> (this IEnumerable<T> source, int i, IEnumerable<T> inserted) {
return source.Take (i).Concat (inserted).Concat (source.Skip (i));
}
/// <summary>
/// 順列を得る
/// O(N!)
/// </summary>
public static IEnumerable<IEnumerable<T>> Perm<T> (this IEnumerable<T> source, int n) {
if (n == 0 || source.IsEmpty () || source.Count () < n) {
return Enumerable.Empty<IEnumerable<T>> ();
} else if (n == 1) {
return source.Select (i => new List<T> { i });
} else {
var nexts = source.Select ((x, i) =>
new { next = source.Take (i).Concat (source.Skip (i + 1)), selected = source.Take (i + 1).Last () });
return nexts.SelectMany (next => Perm (next.next, n - 1).Select (item => item.Prepend (next.selected)));
}
}
/// <summary>
/// シーケンスの隣り合う要素を2引数の関数に適用したシーケンスを得る
/// </summary>
/// <para>O(N)</para>
/// <param name="source">元のシーケンス</param>
/// <param name="func">2引数関数</param>
/// <example>[1,2,3,4].MapAdjacent(f) => [f(1,2), f(2,3), f(3,4)]</example>
public static IEnumerable<TR> MapAdjacent<T1, TR> (this IEnumerable<T1> source, Func<T1, T1, TR> func) {
var list = source.ToList ();
return Enumerable.Range (1, list.Count - 1)
.Select (i => func (list[i], list[i - 1]));
}
/// <summary>
/// 累積項を要素にもつシーケンスを得る(初項は、first)
/// <para>O(N)</para>
/// </summary>
/// <param name="source">元のシーケンス</param>
/// <param name="func">2引数関数f</param>
/// <param name="first">func(first, source[0])のための初項</param>
/// <example> [1,2,3].Scanl1(f,0) => [0, f(0,1), f(f(0,1),2), f(f(f(0,1),2),3)]</example>
public static IEnumerable<TR> Scanl<T, TR> (this IEnumerable<T> source, TR first, Func<TR, T, TR> func) {
var list = source.ToList ();
var result = new List<TR> { first };
foreach (var i in Enumerable.Range (0, source.Count ())) {
result.Add (func (result[i], list[i]));
}
return result;
}
/// <summary>
/// 累積項を要素にもつシーケンスを得る（初項は、source.First()）
/// <para>O(N)</para>
/// </summary>
/// <param name="source">元のシーケンス</param>
/// <param name="func">2引数関数f</param>
/// <example> [1,2,3].Scanl1(f) => [1, f(1,2), f(f(1,2),3)]</example>
public static IEnumerable<T> Scanl1<T> (this IEnumerable<T> source, Func<T, T, T> func) {
var list = source.ToList ();
var result = new List<T> { list[0] };
foreach (var i in Enumerable.Range (1, source.Count () - 1)) {
result.Add (func (result[i - 1], list[i]));
}
return result;
}
/// <summary>
/// 昇順にソートしたインデックスを得る
/// </summary>
/// <para>O(N * log N)</para>
public static IEnumerable<int> SortIndex<T> (this IEnumerable<T> source) {
return source
.Select ((item, i) => new { Item = item, Index = i })
.OrderBy (x => x.Item)
.Select (x => x.Index);
}
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | // The following implementation assumes that the activities
// are already sorted according to their finish time
import java.util.*;
import java.lang.*;
import java.io.*;
class ActivitySelection
{
// Prints a maximum set of activities that can be done by a single
// person, one at a time.
// n --> Total number of activities
// s[] --> An array that contains start time of all activities
// f[] --> An array that contains finish time of all activities
public static void printMaxActivities(int s[], int f[], int n)
{
int i, j;
System.out.print("Following activities are selected : n");
// The first activity always gets selected
i = 0;
System.out.print(i+" ");
// Consider rest of the activities
for (j = 1; j < n; j++)
{
// If this activity has start time greater than or
// equal to the finish time of previously selected
// activity, then select it
if (s[j] >= f[i])
{
System.out.print(j+" ");
i = j;
}
}
}
// driver program to test above function
public static void main(String[] args)
{
int s[] = {1, 3, 0, 5, 8, 5};
int f[] = {2, 4, 6, 7, 9, 9};
int n = s.length;
printMaxActivities(s, f, n);
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
char s[20];
int num[200];
long long sum = 0;
int main() {
int n;
long long k;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
getchar();
scanf("%s", s);
if (s[0] == 'M') {
num[0]++;
} else if (s[0] == 'A') {
num[1]++;
} else if (s[0] == 'R')
num[2]++;
else if (s[0] == 'C')
num[3]++;
else if (s[0] == 'H')
num[4]++;
}
sum += num[0] * num[1] * num[2];
sum += num[0] * num[1] * num[3];
sum += num[0] * num[1] * num[4];
sum += num[0] * num[2] * num[3];
sum += num[0] * num[2] * num[4];
sum += num[0] * num[3] * num[4];
sum += num[1] * num[2] * num[3];
sum += num[1] * num[2] * num[4];
sum += num[1] * num[3] * num[4];
sum += num[2] * num[3] * num[4];
printf("%lld\n", sum);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | fun main(args : Array<String>) {
val l = (1..readLine()!!.toInt()).map { readLine()!![0] }.run {
listOf(count { it == 'M' }, count { it == 'A' }, count { it == 'R' }, count { it == 'C' }, count { it == 'H' }).filter { it != 0 }
}
when {
l.size == 3 -> print(l[0] * l[1] * l[2])
l.size == 4 -> print(l.map { l[0] * l[1] * l[2] * l[3] / it }.sum())
l.size == 5 -> {
print((0..2).map { i ->
(i + 1..3).map { j ->
(j + 1..4).map { k -> l[i] * l[j] * l[k] }.sum()
}.sum()
}.sum())
}
else -> print(0)
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
int N;
cin >> N;
int M, A, R, C, H;
M = 0;
A = 0;
R = 0;
C = 0;
H = 0;
for (int i = 0; i < N; i++) {
string S;
cin >> S;
char s = S[0];
if (s == 'M') M++;
if (s == 'A') A++;
if (s == 'R') R++;
if (s == 'C') C++;
if (s == 'H') H++;
}
vector<int> V = {M, A, R, C, H};
long long ans = 0;
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
ans += V[i] * V[j] * V[k];
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | #pragma warning disable
using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Runtime.Serialization.Formatters.Binary;
using System.Text;
using System.Diagnostics;
using System.Collections;
static class MainClass
{
public static void Main()
{
var n = Console.ReadLine().ToInt32();
var march = new int[5];
for (int i = 0; i < n; i++)
{
var str = Console.ReadLine()[0];
switch (str)
{
case 'M':
march[0]++;
break;
case 'A':
march[1]++;
break;
case 'R':
march[2]++;
break;
case 'C':
march[3]++;
break;
case 'H':
march[4]++;
break;
default:
break;
}
}
var count = 0;
for (int i = 0; i < 5; i++)
{
for (int j = i; j < 5; j++)
{
if (i == j)
continue;
for (int k = j; k < 5; k++)
{
if (i == k || j == k)
continue;
count += march[i] * march[j] * march[k];
}
}
}
Console.WriteLine(count);
Console.ReadLine();
}
#region ライブラリ
#region 二分探索
public static int LowerBound<T>(this IEnumerable<T> ar, int start, int end, T value, IComparer<T> comparer)
{
var arr = ar.ToArray();
int low = start;
int high = end;
int mid;
while (low < high)
{
mid = ((high - low) >> 1) + low;
if (comparer.Compare(arr[mid], value) < 0)
low = mid + 1;
else
high = mid;
}
return low;
}
public static int LowerBound<T>(this IEnumerable<T> arr, T value) where T : IComparable
{
return LowerBound(arr, 0, arr.Count(), value, Comparer<T>.Default);
}
public static int UpperBound<T>(this IEnumerable<T> ar, int start, int end, T value, IComparer<T> comparer)
{
var arr = ar.ToArray();
int low = start;
int high = end;
int mid;
while (low < high)
{
mid = ((high - low) >> 1) + low;
if (comparer.Compare(arr[mid], value) <= 0)
low = mid + 1;
else
high = mid;
}
return low;
}
public static int UpperBound<T>(this IEnumerable<T> arr, T value)
{
return UpperBound(arr, 0, arr.Count(), value, Comparer<T>.Default);
}
#endregion
#region bit全探索
/// <summary>
/// Bit全探索用のライブラリ
/// </summary>
/// <param name="len">作る配列の長さ(lenが0以下だと（ｒｙ)</param>
/// <returns>できた配列</returns>
public static List<List<bool>> CreateBits(int len)
{
return CreateBitsPri(new List<List<bool>>(), len, 0);
}
private static List<List<bool>> CreateBitsPri(List<List<bool>> bits, int len, int count)
{
if (count == 0)
{
var lss = new List<bool>();
lss.Add(true);
bits.Add(lss);
var lx = new List<bool>();
lx.Add(false);
bits.Add(lx);
count++;
return CreateBitsPri(bits, len, count);
}
if (len - count == 0)
return bits;
count++;
var ls = new List<List<bool>>();
foreach (var item in bits)
{
var x = item.ToList();
x.Add(true);
item.Add(false);
ls.Add(item);
ls.Add(x);
}
return CreateBitsPri(ls, len, count);
}
#endregion
#region 文字列処理
public static int ToInt32(this string str)
{
return int.Parse(str);
}
public static List<string> SplittedStringToList(this string str)
{
return str.Split(new[] { ' ' }, StringSplitOptions.RemoveEmptyEntries)
.ToList();
}
public static List<int> SplittedStringToInt32List(this string str)
{
return str.Split(new[] { ' ' }, StringSplitOptions.RemoveEmptyEntries).Select(x => int.Parse(x)).ToList();
}
#endregion
#region エラトステネスの篩
/**
* MIT License
Copyright (c) 2018 gushwell
Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:
The above copyright notice and this permission notice shall be included in all
copies or substantial portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
SOFTWARE.
**/
public class BoundedBoolArray
{
private BitArray _array;
private int _lower;
public BoundedBoolArray(int lower, int upper)
{
_array = new BitArray(upper - lower + 1);
_lower = lower;
}
public bool this[int index]
{
get
{
return _array[index - _lower];
}
set
{
_array[index - _lower] = value;
}
}
}
public static IEnumerable<int> GeneratePrimes()
{
var primes = new List<int>() { 2, 3 };
foreach (var p in primes)
yield return p;
int ix = 0;
while (true)
{
int prime1st = primes[ix];
int prime2nd = primes[++ix];
var lower = prime1st * prime1st;
var upper = prime2nd * prime2nd - 1;
var sieve = new BoundedBoolArray(lower, upper);
foreach (var prime in primes.Take(ix))
{
var start = (int)Math.Ceiling((double)lower / prime) * prime;
for (int index = start; index <= upper; index += prime)
sieve[index] = true;
}
for (int i = lower; i <= upper; i++)
{
if (sieve[i] == false)
{
primes.Add(i);
yield return i;
}
}
}
}
#endregion
#region DeepClone
public static T DeepClone<T>(this T src)
{
using (var memoryStream = new MemoryStream())
{
var binaryFormatter
= new BinaryFormatter();
binaryFormatter.Serialize(memoryStream, src);
memoryStream.Seek(0, SeekOrigin.Begin);
return (T)binaryFormatter.Deserialize(memoryStream);
}
}
#endregion
#region 順列を作るやつ
public static List<List<int>> GeneratePermutations(int n)
{
if (n > 14)
throw new ArgumentOutOfRangeException();
var ls = new List<List<int>>();
ls.Add(new List<int>() { 1 });
for (int i = 0; i < n - 1; i++)
{
var newls = new List<List<int>>();
ls.ForEach(x =>
{
var len = x.Count;
for (int j = 0; j <= len; j++)
{
var item = x.DeepClone();
item.Insert(j, i + 2);
newls.Add(item);
}
});
ls = newls;
}
return ls;
}
#endregion
#endregion
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | def m():
n = int(input())
S = [str(input()) for _ in range(n)]
m = 0
a = 0
r = 0
c = 0
h = 0
x = [0, 0, 0, 0, 0, 0, 1, 1, 1, 2]
y = [1, 1, 1, 2, 2, 3, 2, 2, 3, 3]
z = [2, 3, 4, 3, 4, 3, 4, 4, 4, 4]
for i in range(n):
if S[i][0] == "M":
m += 1
elif S[i][0] == "A":
a += 1
elif S[i][0] == "R":
r += 1
elif S[i][0] == "C":
c += 1
elif S[i][0] == "H":
h += 1
ans = 0
d = [m, a, r, c, h]
for j in range(10):
ans += d[x[j]] * d[y[j]] * d[z[j]]
return ans
print(m())
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;
int main() {
int n;
cin >> n;
vector<int> dp(5);
ll ans = 0;
for (int i = 0; i < (n); i++) {
string s;
cin >> s;
if (s[0] == 'M')
dp[0]++;
else if (s[0] == 'A')
dp[1]++;
else if (s[0] == 'R')
dp[2]++;
else if (s[0] == 'C')
dp[3]++;
else if (s[0] == 'H')
dp[4]++;
}
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += dp[i] * dp[j] * dp[k];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
int c[5] = {};
long long int ans = 0;
cin >> n;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] == 'M')
c[0]++;
else if (s[0] == 'A')
c[1]++;
else if (s[0] == 'R')
c[2]++;
else if (s[0] == 'C')
c[3]++;
else if (s[0] == 'H')
c[4]++;
}
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
for (int k = j + 1; k < n; k++) {
ans += c[i] * c[j] * c[k];
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | using System;
using System.Collections.Generic;
using System.Linq;
using static System.Console;
class Program
{
static void Main()
{
var N = int.Parse(ReadLine());
var S = new string[N];
for (int i = 0; i < N; i++)
{
S[i] = ReadLine();
}
long M = S.Count(x => x.First() == 'M');
long A = S.Count(x => x.First() == 'A');
long R = S.Count(x => x.First() == 'R');
long C = S.Count(x => x.First() == 'C');
long H = S.Count(x => x.First() == 'H');
var L = M * A * R + M * A * C + M * A * H + M * R * C + M * R * H + M * C * H + A * R * C + A * R * H + R * C * H;
WriteLine(L);
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int factorial(int x) {
if (x == 1) return 1;
return x * factorial(x - 1);
}
int main() {
int n;
cin >> n;
int march[5] = {0};
for (int i = 0; i < n; i++) {
string s;
cin >> s;
switch (s[0]) {
case 'M':
march[0]++;
break;
case 'A':
march[1]++;
break;
case 'R':
march[2]++;
break;
case 'C':
march[3]++;
break;
case 'H':
march[4]++;
break;
default:
break;
}
}
int comb[3][10] = {{0, 0, 0, 0, 0, 0, 1, 1, 1, 2},
{1, 1, 1, 2, 2, 3, 2, 2, 3, 3},
{2, 3, 4, 3, 4, 4, 3, 4, 4, 4}};
long long ans = 0;
for (int i = 0; i < 10; i++) {
ans += march[comb[0][i]] * march[comb[1][i]] * march[comb[2][i]];
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
unordered_map<char, int> m;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] != 'M' && s[0] != 'A' && s[0] != 'R' && s[0] != 'C' && s[0] != 'H')
continue;
auto it = m.find(s[0]);
if (it != m.end())
it->second++;
else
m[s[0]] = 1;
}
vector<long long> v;
v.reserve(5);
for (auto&& p : m) v.push_back(p.second);
long long count = 0;
for (int i = 0; i < 3; i++)
for (int j = i + 1; j < 4; j++)
for (int k = j + 1; k < 5; k++) count += v[i] * v[j] * v[k];
cout << count << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MAX = 10000000;
const int MOD = 1000000007;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
fac[0] = fac[1] = 1;
finv[0] = finv[1] = 1;
inv[1] = 1;
for (int i = 2; i < MAX; i++) {
fac[i] = fac[i - 1] * i % MOD;
inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
finv[i] = finv[i - 1] * inv[i] % MOD;
}
}
long long COM(int n, int k) {
if (n < k) return 0;
if (n < 0 || k < 0) return 0;
return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
int main() {
int n;
cin >> n;
vector<int> s(5);
for (int i = 0; i < n; i++) {
string k;
cin >> k;
if (k[0] == 'M')
s[0]++;
else if (k[0] == 'A')
s[1]++;
else if (k[0] == 'R')
s[2]++;
else if (k[0] == 'C')
s[3]++;
else if (k[0] == 'H')
s[4]++;
}
ll total = 0;
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
total += s[i] * s[j] * s[k];
}
}
}
cout << total << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | n=int(input())
s={"M":0,"A":0,"R":0,"C":0,"H":0}
for i in range(n):
v=input()
if v[0] in s:
s[v[0]]+=1
l=s.values()
sum=0
for a in range(5):
for b in range(a+1,5):
for c in range(b+1,5):
sum+=l[a]*l[b]*l[c]
print(sum) |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int N;
int m = 0;
int a = 0;
int r = 0;
int c = 0;
int h = 0;
bool flag = false;
int main(void) {
cin >> N;
for (int i = 0; i < N; i++) {
string t;
cin >> t;
if (t[0] == 'M') m++;
if (t[0] == 'A') a++;
if (t[0] == 'R') r++;
if (t[0] == 'C') c++;
if (t[0] == 'H') h++;
}
long long ans = m * a * r + m * a * c;
ans += m * a * h + m * r * c;
ans += m * r * h + m * c * h;
ans += a * r * c + a * r * h;
ans += a * c * h + r * c * h;
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
string s;
char p;
int m = 0;
int a = 0;
int r = 0;
int c = 0;
int h = 0;
for (int i = 0; i < N; i++) {
cin >> s;
p = s.at(0);
if (p == 'M') {
m++;
} else if (p == 'A') {
a++;
} else if (p == 'R') {
r++;
} else if (p == 'C') {
c++;
} else if (p == 'H') {
h++;
}
}
cout << m * a * r + m * a * c + m * a * h + m * r * c + m * r * h +
m * c * h + a * r * c + a * r * h + a * c * h + r * c * h
<< endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main(void) {
long long N;
cin >> N;
string S;
vector<char> key(N);
int m = 0, a = 0, r = 0, c = 0, h = 0;
for (int i = 0; i < N; i++) {
cin >> S;
key.at(i) = S.at(0);
if (key.at(i) == 'M') {
m++;
}
if (key.at(i) == 'A') {
a++;
}
if (key.at(i) == 'R') {
r++;
}
if (key.at(i) == 'C') {
c++;
}
if (key.at(i) == 'H') {
h++;
}
}
long long ans = 0;
vector<int> count{m, a, r, c, h};
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += count.at(i) * count.at(j) * count.at(k);
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long cnt[5];
int next_combination(int sub) {
int x = sub & -sub, y = sub + x;
return (((sub & ~y) / x) >> 1) | y;
}
int main() {
int n;
cin >> n;
string name[n];
for (int i = 0; i < (n); ++i) cin >> name[i];
for (int i = 0; i < (n); ++i) {
char x = name[i][0];
if (x == 'M')
++cnt[0];
else if (x == 'A')
++cnt[1];
else if (x == 'R')
++cnt[2];
else if (x == 'C')
++cnt[3];
else if (x == 'H')
++cnt[4];
}
long long ans = 0;
int k = 3;
for (int bit = (1 << k) - 1; bit < (1 << n); bit = next_combination(bit)) {
long long tmp = 1;
for (int i = 0; i < 5; ++i) {
if ((bit >> i) & 1) tmp *= cnt[i];
}
ans += tmp;
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("Ofast")
constexpr ll INF = 1e18;
constexpr int inf = 1e9;
constexpr double INFD = 1e100;
constexpr ll mod = 1000000007;
constexpr ll mod2 = 998244353;
const double PI = 3.1415926535897932384626433832795028841971;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
template <class T>
inline bool chmax(T& a, T b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
if (a > b) {
a = b;
return 1;
}
return 0;
}
struct mint {
ll x;
mint(ll x = 0) : x((x % mod + mod) % mod) {}
mint& operator+=(const mint a) {
if ((x += a.x) >= mod) x -= mod;
return *this;
}
mint& operator-=(const mint a) {
if ((x += mod - a.x) >= mod) x -= mod;
return *this;
}
mint& operator*=(const mint a) {
(x *= a.x) %= mod;
return *this;
}
mint operator+(const mint a) const {
mint res(*this);
return res += a;
}
mint operator-(const mint a) const {
mint res(*this);
return res -= a;
}
mint operator*(const mint a) const {
mint res(*this);
return res *= a;
}
mint pow(ll t) const {
if (!t) return 1;
mint a = pow(t >> 1);
a *= a;
if (t & 1) a *= *this;
return a;
}
mint inv() const { return pow(mod - 2); }
mint& operator/=(const mint a) { return (*this) *= a.inv(); }
mint operator/(const mint a) const {
mint res(*this);
return res /= a;
}
};
struct combination {
vector<mint> fact, ifact;
combination(int n) : fact(n + 1), ifact(n + 1) {
assert(n < mod);
fact[0] = 1;
for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
ifact[n] = fact[n].inv();
for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
}
mint operator()(int n, int k) {
if (k < 0 || k > n) return 0;
return fact[n] * ifact[k] * ifact[n - k];
}
};
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int N;
cin >> N;
vector<string> S(N);
for (int i = 0; i < N; i++) {
cin >> S[i];
}
vector<int> cnt(5, 0);
string str = "MARCH";
for (int i = 0; i < N; i++) {
for (int j = 0; j < 5; j++) {
if (str[j] == S[i].front()) {
cnt[j]++;
}
}
}
int ans = 0;
for (int bit = 0; bit < (1 << 5); bit++) {
if (__builtin_popcount(bit) == 3) {
int temp = 1;
for (int i = 0; i < 5; i++) {
if (bit & (1 << i)) {
temp *= cnt[i];
}
}
ans += temp;
}
}
cout << ans << '\n';
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.Scanner;
import java.util.ArrayList;
public class Main {
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
String name[] = new String[n];
for(int t = 0;t<n;t++) {
name[t] = scanner.next();
}
int x = 0;
Long amnt[] = {0L,0L,0L,0L,0L};
String moji[] = {"M","A","R","C","H"};
int s;
for(x = 0;x < n;x++){
for(s = 0;s<5;s++){
if(moji[s].equals(name[x].charAt(0)))amnt[s]++;
}
}
Long ans = amnt[0]*(amnt[1]*(amnt[2]+amnt[3]+amnt[4])+amnt[2]*(amnt[3]+amnt[4])+amnt[3]*amnt[4])+amnt[1]*(amnt[2]*amnt[3]+amnt[2]*amnt[4]+amnt[3]*amnt[4])+amnt[2]*amnt[3]*amnt[4];
System.out.println(ans);
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using ll = long long;
using namespace std;
const int INF = 1e9;
int main() {
int N;
cin >> N;
int cnt[5] = {0};
string T = "MARCH";
for (int i = 0; i < N; ++i) {
string S;
cin >> S;
for (int i = 0; i < N; ++i)
if (S[0] == T[i]) {
++cnt[i];
break;
}
}
ll ans = 0;
for (int i = 0; i < N; ++i) {
for (int j = i; j < N; ++j) {
for (int k = j; k < N; ++k) {
if (i == j || j == k) continue;
ans += cnt[i] * cnt[j] * cnt[k];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.Scanner;
class Main{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = 0;
int a = 0;
int r = 0;
int c = 0;
int h = 0;
for(int i = 0; i < n; i++){
String str = sc.next().substring(0,1);
if(str.equals("M")){
m++;
}
if(str.equals("A")){
a++;
}
if(str.equals("R")){
r++;
}
if(str.equals("C")){
c++;
}
if(str.equals("H")){
h++;
}
}
long ans = m*a*r + m*a*c + m*a*h + m*r*c + m*r*h + m*c*h + a*r*c + a*r*h + a*c*h + r*c*h;
System.out.println(ans);
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.*;
public class Main {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int n = Integer.parseInt(sc.next());
String[] s = new String[n];
long[] march = new long[5];
for(int i=0;i<n;i++){
s[i] = sc.next();
if(s[i].charAt(0) == 'M') march[0]++;
else if(s[i].charAt(0) == 'A') march[1]++;
else if(s[i].charAt(0) == 'R') march[2]++;
else if(s[i].charAt(0) == 'C') march[3]++;
else if(s[i].charAt(0) == 'H') march[4]++;
}
long ans = 0;
for(int i=0; i<n-2; i++) {
for(int j=i+1; j<n-1; j++) {
for(int k=j+1; k<n; k++) {
ans += march[i]*march[j]*march[k];
}
}
}
System.out.println(ans);
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <class T>
bool chmax(T &a, const T &b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
if (b < a) {
a = b;
return 1;
}
return 0;
}
int N, cnt[5];
char check[5] = {'M', 'A', 'R', 'C', 'H'};
int main() {
cin >> N;
string S[N];
for (int i = 0; i < (int)(N); i++) {
cin >> S[i];
for (int j = 0; j < (int)(5); j++) {
if (S[i][0] == check[j]) {
cnt[j]++;
}
}
}
long long ans = 0;
for (int i = 0; i < (int)(5); i++) {
for (int j = (int)(i + 1); j < (int)(5); j++) {
for (int k = (int)(j + 1); k < (int)(5); k++) {
ans += cnt[i] * cnt[j] * cnt[k];
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("O0")
using namespace std;
template <typename LenType = int>
LenType readln(char* buf, LenType max) {
fgets(buf, max + 1, stdin);
buf[strcspn(buf, "\n")] = 0;
}
template <typename T>
int orderDesc(const void* p1, const void* p2) {
const T& a = *static_cast<const T*>(p1);
const T& b = *static_cast<const T*>(p2);
return (a < b);
}
template <typename T>
int orderAsc(const void* p1, const void* p2) {
const T& a = *static_cast<const T*>(p1);
const T& b = *static_cast<const T*>(p2);
return (a > b);
}
int N;
char S[10001][11];
int hashfunc(char c) {
switch (c) {
case 'M':
return 0;
case 'A':
return 1;
case 'R':
return 2;
case 'C':
return 3;
case 'H':
return 4;
}
return -1;
}
void solve() {
int counts[5] = {0, 0, 0, 0, 0};
int hashcode;
for (int i = 1; i <= N; ++i) {
hashcode = hashfunc(S[i][0]);
if (hashcode != -1) ++counts[hashcode];
}
long long res = 0;
for (int i = 0; i <= 2; ++i) {
for (int j = i + 1; j <= 3; ++j) {
for (int x = j + 1; x <= 4; ++x) {
res += counts[i] * counts[j] * counts[x];
}
}
}
cout << res << endl;
}
int main() {
scanf("%d\n", &N);
for (int i = 1; i <= N && !feof(stdin); ++i) {
readln(S[i], 10);
}
solve();
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
int main() {
int n;
cin >> n;
vector<int> cnt(26);
for (int i = 0; i < (n); i++) {
string s;
cin >> s;
cnt[s[0]]++;
}
ll sum = 0;
sum += cnt['M'] * cnt['A'] * cnt['R'];
sum += cnt['M'] * cnt['A'] * cnt['C'];
sum += cnt['M'] * cnt['A'] * cnt['H'];
sum += cnt['M'] * cnt['R'] * cnt['C'];
sum += cnt['M'] * cnt['R'] * cnt['H'];
sum += cnt['M'] * cnt['C'] * cnt['H'];
sum += cnt['A'] * cnt['R'] * cnt['C'];
sum += cnt['A'] * cnt['R'] * cnt['H'];
sum += cnt['A'] * cnt['C'] * cnt['H'];
sum += cnt['R'] * cnt['C'] * cnt['H'];
cout << sum << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
int N;
scanf("%d", &N);
long long cnt[5] = {0};
for (int i = 0; i < N; i++) {
scanf(" %c%*[^\n]", &c);
// printf("%c\n", c);
switch (c) {
case 'M':
cnt[0]++;
break;
case 'A':
cnt[1]++;
break;
case 'R':
cnt[2]++;
break;
case 'C':
cnt[3]++;
break;
case 'H':
cnt[4]++;
break;
default:
break;
}
}
long long result = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
result += cnt[i] * cnt[j] * cnt[k];
// printf("%d %d %d %lld \n", i, j, k, result);
}
}
}
printf("%lld", result);
return 0;
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N, i, a[5], j, k;
cin >> N;
string s;
for (i = 0; i < 5; i++) {
a[i] = 0;
}
for (i = 0; i < N; i++) {
cin >> s;
if (s.substr(0, 1) == "M") {
a[0] += 1;
} else if (s.substr(0, 1) == "A") {
a[1] += 1;
} else if (s.substr(0, 1) == "R") {
a[2] += 1;
} else if (s.substr(0, 1) == "C") {
a[3] += 1;
} else if (s.substr(0, 1) == "H") {
a[4] += 1;
}
}
long long n = 0;
for (i = 0; i < 3; i++) {
for (j = i + 1; j < 4; j++) {
for (k = j + 1; k < 5; k++) {
n += a[i] * a[j] * a[k];
}
}
}
cout << n;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
enum Index { M = 0, A, R, C, H, Max };
int count[5] = {0};
int P[] = {0, 0, 0, 0, 0, 0, 1, 1, 1, 2};
int Q[] = {1, 1, 1, 2, 2, 3, 2, 2, 3, 3};
int K[] = {2, 3, 4, 3, 4, 4, 3, 4, 4, 4};
int N;
scanf(" %d", &N);
for (int i = 0; i < N; i++) {
string s;
cin >> s;
switch (s[0]) {
case 'M': {
count[Index::M]++;
break;
}
case 'A': {
count[Index::A]++;
break;
}
case 'R': {
count[Index::R]++;
break;
}
case 'C': {
count[Index::C]++;
break;
}
case 'H': {
count[Index::H]++;
break;
}
}
}
long long res = 0;
for (int i = 0; i < 10; i++) {
res += count[P[i]] * count[Q[i]] * count[K[i]];
}
printf("%d\n", res);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<string> s(n);
for (int i = 0; i < n; i++) cin >> s[i];
vector<int> count(5, 0);
for (int i = 0; i < n; i++) {
if (s[i][0] == 'M')
count[0]++;
else if (s[i][0] == 'A')
count[1]++;
else if (s[i][0] == 'R')
count[2]++;
else if (s[i][0] == 'C')
count[3]++;
else if (s[i][0] == 'H')
count[4]++;
}
long long int ans = 0;
for (int bit = 0; bit < (1 << 5); bit++) {
int coans = 1;
int num = 0;
for (int i = 0; i < 5; i++) {
if (bit & (1 << i)) {
coans *= count[i];
num++;
}
}
if (num == 3) ans += coans;
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
int c[5] = {};
cin >> N;
char **names = new char *[N];
for (int i = 0; i < N; i++) {
names[i] = new char[10];
cin >> names[i];
if (names[i][0] == 'M') c[0]++;
if (names[i][0] == 'A') c[1]++;
if (names[i][0] == 'R') c[2]++;
if (names[i][0] == 'C') c[3]++;
if (names[i][0] == 'H') c[4]++;
}
long long combination = 0;
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
combination += c[i] * c[j] * c[k];
}
}
}
cout << combination << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int q[6];
int main() {
int n, ans = 0;
char ch[11];
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%s", ch);
if (ch[0] == 'M') q[1]++;
if (ch[0] == 'A') q[2]++;
if (ch[0] == 'R') q[3]++;
if (ch[0] == 'C') q[4]++;
if (ch[0] == 'H') q[5]++;
}
for (int i = 1; i <= 3; i++) {
for (int j = i + 1; j <= 4; j++) {
for (int k = j + 1; k <= 5; k++) {
if (q[i] == 0 || q[j] == 0 || q[k] == 0)
;
else
ans = ans + q[i] * q[j] * q[k];
}
}
}
printf("%d", ans);
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | #include <bits/stdc++.h>
int main() {
unsigned long ans = 0;
int i, N, M = 0, A = 0, R = 0, C = 0, H = 0;
char name[20];
scanf("%d", &N);
for (i = 0; i < N; i++) {
scanf("%s", name);
if (name[0] == 'M') M++;
if (name[0] == 'A') A++;
if (name[0] == 'R') R++;
if (name[0] == 'C') C++;
if (name[0] == 'H') H++;
}
ans = M * A * (R + C + H) + M * R * (C + H) + M * C * H + A * R * (C + H) +
(A + R) * C * H;
printf("%lu\n", ans);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <class T>
void _R(T &x) {
cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%" PRId64, &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
_R(head);
R(tail...);
}
template <class T>
void _W(const T &x) {
cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%" PRId64, x); }
void _W(const double &x) { printf("%.16f\n", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T>
void _W(const vector<T> &x) {
for (auto i = x.begin(); i != x.end(); _W(*i++))
if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
_W(head);
putchar(sizeof...(tail) ? ' ' : '\n');
W(tail...);
}
template <class T>
inline bool chmax(T &a, const T &b) {
return b > a ? a = b, true : false;
}
template <class T>
inline bool chmin(T &a, const T &b) {
return b < a ? a = b, true : false;
}
template <class T, class F = less<T>>
void sort_uniq(vector<T> &v, F f = F()) {
sort(begin(v), end(v), f);
v.resize(unique(begin(v), end(v)) - begin(v));
}
template <class T>
using MaxHeap = priority_queue<T>;
template <class T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
int n;
int64_t a[5];
int main() {
R(n);
for (int i = 0; i < int(n); i++) {
char s[15];
R(s);
if (s[0] == 'M') a[0]++;
if (s[0] == 'A') a[1]++;
if (s[0] == 'R') a[2]++;
if (s[0] == 'C') a[3]++;
if (s[0] == 'H') a[4]++;
}
int64_t ans = 0;
for (int i = 0; i < int(5); i++) {
for (int j = (i + 1); j <= int(5); j++) {
for (int k = (j + 1); k <= int(5); k++) {
ans += a[i] * a[j] * a[k];
}
}
}
W(ans);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
int n;
cin >> n;
int march[20];
for (int i = 0; i < 20; i++) {
march[i] = 0;
}
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] == 'M')
march[0]++;
else if (s[0] == 'A')
march[1]++;
else if (s[0] == 'R')
march[2]++;
else if (s[0] == 'C')
march[3]++;
else if (s[0] == 'H')
march[4]++;
}
ll ans = 0;
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
ans += march[i] * march[j] * march[k];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | egrep '^M|^A|^R|^C|^H'|cut -c-1|tr MARCH 1-5|awk '{a[$0]++}END{b=0;for(i=1;i<=5;i++)for(j=i+1;j<=5;j++)for(k=j+1;k<=5;k++)b+=a[i]*a[j]*a[k];print b}' |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
int main(void) {
long long n, i, j, k;
char s[30];
int a[5];
for (i = 0; i < 5; ++i) a[i] = 0;
scanf("%lld", &n);
for (i = 0; i < n; ++i) {
scanf("%s", s);
if (s[0] == 'M')
a[0]++;
else if (s[0] == 'A')
a[1]++;
else if (s[0] == 'R')
a[2]++;
else if (s[0] == 'C')
a[3]++;
else if (s[0] == 'H')
a[4]++;
}
long long ans = 0;
for (i = 0; i < 5; ++i)
for (j = i + 1; j < 5; ++j)
for (k = j + 1; k < 5; ++k) ans += a[i] * a[j] * a[k];
printf("%lld", ans);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int mod = 1000000007;
char s[22];
int main() {
int T, n, k;
while (~scanf("%d", &n)) {
char rec[256] = {0};
for (int i = 0; i < n; i++) {
scanf("%s", s);
rec[s[0]]++;
}
char cs[] = "MARCH";
long long ans = 0;
for (int i = 0; i < 32; i++) {
int cnt = 0;
for (int j = 0; j < 5; j++) cnt += i >> j & 1;
if (cnt == 3) {
long long t = 1;
for (int j = 0; j < 5; j++)
if (i >> j & 1) t *= rec[cs[j]];
ans += t;
}
}
cout << ans << endl;
}
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long INF = 1000000000;
int d[5];
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
int n;
cin >> n;
for (int i = 0; i < (n); ++i) {
string s;
cin >> s;
if (s[0] == 'M') {
d[0]++;
} else if (s[0] == 'A') {
d[1]++;
} else if (s[0] == 'R') {
d[2]++;
} else if (s[0] == 'C') {
d[3]++;
} else if (s[0] == 'H') {
d[4]++;
}
}
int ans = 0;
for (int i = 0; i < (5); ++i) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += d[i] * d[j] * d[k];
}
}
}
cout << ans << "\n";
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
vector<string> S(N);
for (int i = 0; i < N; i++) {
cin >> S.at(i);
}
int m = 0, a = 0, r = 0, c = 0, h = 0;
for (int i = 0; i < N; i++) {
char I = S.at(i).at(0);
if (I == 'M') m++;
if (I == 'A') a++;
if (I == 'R') r++;
if (I == 'C') c++;
if (I == 'H') h++;
}
long long ans = 0;
ans = m * a * r + m * a * c + m * a * h + m * r * c + m * r * h + m * c * h +
a * r * c + a * r * h + a * c * h + r * c * h;
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long m = 0, a = 0, r = 0, c = 0, h = 0;
signed main() {
long long N;
cin >> N;
for (long long i = 0; i < N; i++) {
string s;
cin >> s;
switch (s[0]) {
case 'M':
m++;
break;
case 'A':
a++;
break;
case 'R':
r++;
break;
case 'C':
c++;
break;
case 'H':
h++;
break;
}
}
long long result = m * a * r + m * a * c + m * a * h + m * r * c + m * r * h +
m * c * h + a * r * c + a * r * h + r * c * h;
cout << result << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | #[macro_use]
mod input {
use std;
use std::io;
const SPLIT_DELIMITER: char = ' ';
#[macro_export]
#[allow(unused_macros)]
macro_rules! input {
( $($x:expr ),*) => {
{
let temp_str = input_line_str();
let mut split_result_iter = temp_str.split_whitespace();
$(
let buf_split_result = split_result_iter.next();
let buf_split_result = buf_split_result.unwrap();
($x) = buf_split_result.parse().unwrap();
)*
}
};
}
#[allow(dead_code)]
pub fn input_line_str() -> String {
let mut s = String::new();
io::stdin().read_line(&mut s).unwrap();
s.trim().to_string()
}
#[allow(dead_code)]
pub fn p<T>(t: T)
where
T: std::fmt::Display,
{
println!("{}", t);
}
#[allow(dead_code)]
pub fn input_vector2d<T>(line: usize) -> Vec<Vec<T>>
where
T: std::str::FromStr,
{
let mut v: Vec<Vec<T>> = Vec::new();
for _ in 0..line {
let vec_line = input_vector();
v.push(vec_line);
}
v
}
#[allow(dead_code)]
pub fn input_vector<T>() -> Vec<T>
where
T: std::str::FromStr,
{
let mut v: Vec<T> = Vec::new();
let s = input_line_str();
let split_result = s.split(SPLIT_DELIMITER);
for z in split_result {
let buf = match z.parse() {
Ok(r) => r,
Err(_) => panic!("Parse Error"),
};
v.push(buf);
}
v
}
#[allow(dead_code)]
pub fn str2vec(s: &str) -> Vec<char> {
let mut v: Vec<char> = Vec::new();
for c in s.chars() {
v.push(c);
}
v
}
}
use input::*;
use std::collections::HashMap;
fn main() {
let n: usize;
input!(n);
let mut map = HashMap::new();
for _ in 0..n {
let s = str2vec(&input_line_str());
if !(s[0] == 'M' || s[0] == 'A' || s[0] == 'R' || s[0] == 'C' || s[0] == 'H') {
continue;
}
let po = map.entry(s[0]).or_insert(0);
*po += 1;
}
let mut ans = 0;
let mut v = vec![];
for (_, x) in map {
v.push(x);
}
for i in 0..v.len() {
for j in i + 1..v.len() {
for k in j + 1..v.len() {
ans += v[i] * v[j] * v[k];
}
}
}
p(ans);
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N, i, j = 0, k = 0, l = 0, m = 0, n = 0;
char S[100];
cin >> N;
for (i = 0; i < N; i++) {
cin >> S[i];
if (S[i] == 'M') j++;
if (S[i] == 'A') k++;
if (S[i] == 'R') l++;
if (S[i] == 'C') m++;
if (S[i] == 'H') n++;
}
i = j + k + l + m + n;
if (i >= 3) cout << i * (i - 1) * (i - 2) / 6 << endl;
if (i < 3) cout << 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using PII = pair<int, int>;
using PLL = pair<ll, ll>;
const int INF = numeric_limits<int>::max();
const ll INFL = numeric_limits<ll>::max();
long long mod(long long val, long long m) {
long long res = val % m;
if (res < 0) res += m;
return res;
}
long long gcd(ll a, ll b) {
if (a % b == 0) {
return b;
} else {
return gcd(b, a % b);
}
}
long long lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll factorial(ll n) {
ll res = 1;
for (ll i = 1; i <= n; i++) {
res *= i;
}
return res;
}
int main() {
int n;
cin >> n;
vector<int> s(5, 0);
int p[10] = {0, 0, 0, 0, 0, 0, 1, 1, 1, 2};
int q[10] = {1, 1, 1, 2, 2, 3, 2, 2, 3, 3};
int r[10] = {2, 3, 4, 3, 4, 4, 3, 4, 4, 4};
for (int i = 0; i < (n); i++) {
string tmp;
cin >> tmp;
if (tmp[0] == 'M') {
s[0]++;
}
if (tmp[0] == 'A') {
s[1]++;
}
if (tmp[0] == 'R') {
s[2]++;
}
if (tmp[0] == 'C') {
s[3]++;
}
if (tmp[0] == 'H') {
s[4]++;
}
}
ll ans = 0;
for (int i = 0; i < (10); i++) {
ans += s[p[i]] * s[q[i]] * s[r[i]];
}
printf("% lld\n", ans);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, M = 0, A = 0, R = 0, C = 0, H = 0, i;
long long s = 0LL, a[10];
char L[100009][100];
cin >> n;
for (i = 0; i < n; i++) {
cin >> L[i];
}
for (i = 0; i < n; i++) {
switch (L[i][0]) {
case 'M':
M++;
break;
case 'A':
A++;
break;
case 'R':
R++;
break;
case 'C':
C++;
break;
case 'H':
H++;
break;
default:
break;
}
}
a[0] = M * A * R;
a[1] = M * A * C;
a[2] = M * A * H;
a[3] = M * R * C;
a[4] = M * R * H;
a[5] = M * C * H;
a[6] = A * R * C;
a[7] = A * R * H;
a[8] = A * C * H;
a[9] = R * C * H;
for (i = 0; i < 10; i++) {
s += a[i];
}
cout << s << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | using System;
using System.Linq;
namespace ConsoleApp1
{
class Program
{
static void Main(string[] a_asArgs)
{
int l_iN = Console.ReadLine().Split(' ').Select(a => int.Parse(a)).ToArray()[0];
char[] l_acHead = new char[] { 'M', 'A', 'R', 'C', 'H' };
int[] l_aiCount = new int[5];
for (int l_iIndex = 0; l_iIndex < l_iN; l_iIndex++)
{
string l_sName = Console.ReadLine();
int l_iPos = Array.IndexOf<char>(l_acHead, l_sName[0]);
if (l_iPos > -1)
{
l_aiCount[l_iPos]++;
}
}
int[][] l_aaiCombination = new int[][] {
new int[] {0,1,2},
new int[] {0,1,3},
new int[] {0,1,4},
new int[] {0,2,3},
new int[] {0,2,4},
new int[] {0,3,4},
new int[] {1,2,3},
new int[] {1,2,4},
new int[] {1,3,4},
new int[] {2,3,4},
};
long l_iResult = 0;
for (int l_iBuf = 0; l_iBuf < l_aaiCombination.Length; l_iBuf++)
{
l_iResult += l_aiCount[l_aaiCombination[l_iBuf][0]] * l_aiCount[l_aaiCombination[l_iBuf][1]] * l_aiCount[l_aaiCombination[l_iBuf][2]];
}
Console.WriteLine(l_iResult);
}
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int P[10] = {0, 0, 0, 0, 0, 0, 1, 1, 1, 2};
int Q[10] = {1, 1, 1, 2, 2, 3, 2, 2, 3, 3};
int R[10] = {2, 3, 4, 3, 4, 4, 3, 4, 4, 4};
int main() {
int n;
char name[] = {'M', 'A', 'R', 'C', 'H'};
int cnt[5] = {};
cin >> n;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
for (int j = 0; j < 5; j++) {
if (s[0] == name[j]) {
cnt[j]++;
}
}
}
int check = 0;
for (int i = 0; i < 5; i++) {
if (!cnt[i]) check++;
}
if (n - check >= 3) {
long long ans = 0;
for (int i = 0; i < 10; i++) {
ans += cnt[P[i]] * cnt[Q[i]] * cnt[R[i]];
}
cout << ans << endl;
} else
cout << 0 << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b != 0 ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
int main() {
long long n;
cin >> n;
long long cnt[5] = {0};
string s, march = "MARCH";
for (long long i = 0; i < (long long)(n); i++) {
cin >> s;
for (long long j = 0; j < (long long)(5); j++) {
if (s[0] = march[j]) cnt[j]++;
}
}
long long ans = 0;
for (long long i = 0; i < (long long)(5); i++) {
for (long long j = i + 1; j < 5; j++) {
for (long long k = j + 1; j < 5; j++) {
ans += cnt[i] * cnt[j] * cnt[k];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
string S;
int m = 0, a = 0, r = 0, c = 0, h = 0;
for (int i = 0; i < N; i++) {
cin >> S;
if (S.at(0) == 'M') {
m++;
}
if (S.at(0) == 'A') {
a++;
}
if (S.at(0) == 'R') {
r++;
}
if (S.at(0) == 'C') {
c++;
}
if (S.at(0) == 'H') {
h++;
}
}
cout << m * a * r + m * a * c + m * a * h + m * r * c + m * r * h +
m * c * h + a * r * c + a * r * h + a * c * h + r * c * h
<< endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
void chmin(int &a, int b) {
if (a > b) a = b;
}
int main() {
int N;
cin >> N;
vector<string> S(N);
for (int i = 0; i < N; i++) cin >> S[i];
map<char, long long> ma;
char list[5] = {'M', 'A', 'R', 'C', 'H'};
for (int i = 0; i < 5; i++) ma[list[i]] = 0;
for (int i = 0; i < N; i++) ma[S[i].front()]++;
for (int i = 0; i < 5; i++) {
cout << list[i] << ": " << ma[list[i]] << endl;
}
long long res;
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
res += ma[list[i]] + ma[list[j]] + ma[list[k]];
}
}
}
cout << res << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
try (Scanner sc = new Scanner(System.in);) {
solve(sc);
}
}
public static void solve(Scanner sc) {
int[] nums = new int[5];
int n = sc.nextInt();
for (int i = 0; i < n; i++) {
String name = sc.next();
switch (name.charAt(0)) {
case 'M':
nums[0]++;
break;
case 'A':
nums[1]++;
break;
case 'R':
nums[2]++;
break;
case 'C':
nums[3]++;
break;
case 'H':
nums[4]++;
break;
}
}
long ans = 0l;
for (int i = 0; i < n - 2; i++) {
for (int j = i + 1; j < n - 1; j++) {
for (int k = j + 1; k < n; k++) {
ans += nums[i] * nums[j] * nums[k];
}
}
}
System.out.println(ans);
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <typename LenType = int>
LenType readln(char* buf, LenType max) {
LenType len = 0;
while ((max-- > 0) && ((buf[len] = fgetc(stdin)) != '\n') &&
(buf[len] != EOF))
++len;
buf[len] = 0;
return len;
}
template <typename T>
int orderDesc(const void* p1, const void* p2) {
const T& a = *static_cast<const T*>(p1);
const T& b = *static_cast<const T*>(p2);
return (a < b);
}
template <typename T>
int orderAsc(const void* p1, const void* p2) {
const T& a = *static_cast<const T*>(p1);
const T& b = *static_cast<const T*>(p2);
return (a > b);
}
int N;
char S[10000][11];
int hashfunc(char c) {
switch (c) {
case 'M':
return 0;
case 'A':
return 1;
case 'R':
return 2;
case 'C':
return 3;
case 'H':
return 4;
}
return -1;
}
void solve() {
int counts[5] = {};
int hashcode;
for (int i = 1; i <= N; ++i) {
hashcode = hashfunc(S[i][0]);
if (hashcode != -1) ++counts[hashcode];
}
long long res = 0;
for (int i = 0; i <= 3; ++i) {
for (int j = i + 1; j <= 4; ++j) {
for (int x = j + 1; x <= 5; ++x) {
res += counts[i] * counts[j] * counts[x];
}
}
}
cout << res << endl;
}
int main() {
scanf("%d\n", &N);
for (int i = 1; i <= N && !feof(stdin); ++i) {
readln(S[i], 11);
}
solve();
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
void chmin(int &a, int b) {
if (a > b) a = b;
}
int main() {
int N;
cin >> N;
vector<string> S(N);
for (int i = 0; i < N; i++) cin >> S[i];
map<char, int> ma;
char list[5] = {'M', 'A', 'R', 'C', 'H'};
for (int i = 0; i < 5; i++) ma[list[i]] = 0;
for (int i = 0; i < N; i++) ma[S[i].front()]++;
long long res = 0;
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
res += ma[list[i]] * ma[list[j]] * ma[list[k]];
}
}
}
cout << res << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chmin(T &a, const T &b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template <typename T>
bool chmax(T &a, const T &b) {
if (a < b) {
a = b;
return true;
}
return false;
}
const long long INF = 1000000000000000000;
const long long MOD = 1e9 + 7;
const int MAX = 510000;
long long fac[MAX], finv[MAX], inv[MAX];
void ComInit() {
fac[0] = fac[1] = 1;
finv[0] = finv[1] = 1;
inv[1] = 1;
for (int i = 2; i < MAX; i++) {
fac[i] = fac[i - 1] * i % MOD;
inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
finv[i] = finv[i - 1] * inv[i] % MOD;
}
}
long long Com(int n, int k) {
if (n < k) return 0;
if (n < 0 || k < 0) return 0;
return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
int main() {
int n;
cin >> n;
map<char, int> mp;
for (int i = 0; i < (n); i++) {
string s;
cin >> s;
char ss = s[0];
if (ss == 'M' || ss == 'A' || ss == 'R' || ss == 'C' || ss == 'H') {
mp[ss]++;
}
}
ComInit();
long long ans = 0;
string sr = "MARCH";
for (int i = int(0); i < int(5); ++i) {
for (int j = int(i + 1); j < int(5); ++j) {
for (int k = int(j + 1); k < int(5); ++k) {
ans += mp[sr[i]] * mp[sr[j]] * mp[sr[k]];
}
}
}
cout << ans << endl;
system("pause");
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | import re
import itertools
n = int(input())
ls = [input() for _ in range(n)]
new_ls = []
ans = 0
for name in ls:
if re.match(r"[MARCH]", name):
new_ls.append(name[:1])
comb_ls = list(itertools.combinations(new_ls, 3))
if new_ls:
for comb in comb_ls:
s = set(comb)
if len(s) == 3:
ans += 1
print(ans)
else:
print(0)
quit()
|
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