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stringlengths 2
88
| description
stringlengths 31
8.62k
| public_tests
dict | private_tests
dict | solution_type
stringclasses 2
values | programming_language
stringclasses 5
values | solution
stringlengths 1
983k
|
|---|---|---|---|---|---|---|
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 double pie = acos(-1.0);
long long gcd(long long a, long long b) {
if (a % b == 0) {
return (b);
} else {
return (gcd(b, a % b));
}
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
bool is_prime(const unsigned n) {
switch (n) {
case 0:
case 1:
return false;
case 2:
case 3:
return true;
}
if (n % 2 == 0 || n % 3 == 0) return false;
for (unsigned i = 5; i * i <= n; i += 6) {
if (n % i == 0) return false;
if (n % (i + 2) == 0) return false;
}
return true;
}
void solve() {
long long n;
cin >> n;
long long m[] = {0, 0, 0, 0, 0};
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] == 'M') m[0]++;
if (s[0] == 'A') m[1]++;
if (s[0] == 'R') m[2]++;
if (s[0] == 'C') m[3]++;
if (s[0] == 'H') m[4]++;
}
long long ans = 0;
for (int i = 2; i < 5; i++) ans += m[0] * m[1] * m[i];
for (int i = 3; i < 5; i++) ans += m[0] * m[2] * m[i];
ans += m[0] * m[3] * m[4];
for (int i = 3; i < 5; i++) ans += m[1] * m[2] * m[i];
ans += m[2] * m[3] * m[4];
cout << ans << endl;
}
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
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;
int main() {
int n;
cin >> n;
vector<int> march(5);
int m = 0, a = 0, r = 0, c = 0, h = 0;
string s;
for (int i = 0; i < (n); ++i) {
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++;
}
int cnt = 0, sum = 0;
for (int i = 0; i < (5); ++i) {
if (march[i] != 0) {
cnt++;
sum += march[i] - 1;
}
}
march[0] = m, march[1] = a, march[2] = r, march[3] = c, march[4] = h;
long long ans = 0;
for (int i = 0; i < (5); ++i) {
for (int j = int(i + 1); j < int(5); ++j) {
for (int k = int(j + 1); k < int(5); ++k)
ans += march[i] * march[j] * march[k];
}
}
cout << 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 Nmax = 200010;
int nCr(int n, int r) {
int ans = 1;
for (int i = n; i > n - r; --i) {
ans = ans * i;
}
for (int i = 1; i < r + 1; ++i) {
ans = ans / i;
}
return ans;
}
int main() {
long long N;
cin >> N;
string S[Nmax];
char c[Nmax];
int cnt[5] = {}, count = 0;
for (int i = 0; i < N; i++) {
cin >> S[i];
string str(S[i]);
c[i] = str[0];
if (c[i] == 'M') {
cnt[0]++;
count++;
}
if (c[i] == 'A') {
cnt[1]++;
count++;
}
if (c[i] == 'R') {
cnt[2]++;
count++;
}
if (c[i] == 'C') {
cnt[3]++;
count++;
}
if (c[i] == 'H') {
cnt[4]++;
count++;
}
}
if (count <= 2)
cout << "0" << endl;
else {
count = 0;
int count1 = 0;
int n = 1, N = 0;
for (int i = 0; i < 5; i++)
if (cnt[i] >= 1) {
n *= cnt[i];
N += cnt[i];
count++;
if (cnt[i] == 1) count1++;
}
if (count1 >= 3)
cout << n * nCr(count, 3) + (1 - n) * nCr(count1, 3) << endl;
else
cout << n + nCr(count, 3) << 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;
map<char, int> S;
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') {
S[s[0]]++;
}
}
long long answer = 0;
answer += S['M'] * S['A'] * S['R'];
answer += S['M'] * S['A'] * S['C'];
answer += S['M'] * S['A'] * S['H'];
answer += S['M'] * S['R'] * S['C'];
answer += S['M'] * S['R'] * S['H'];
answer += S['M'] * S['C'] * S['H'];
answer += S['A'] * S['R'] * S['C'];
answer += S['A'] * S['R'] * S['H'];
answer += S['A'] * S['C'] * S['H'];
answer += S['R'] * S['C'] * S['H'];
cout << answer << 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 | #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.Numerics;
static class MainClass
{
public static void Main()
{
var n = Console.ReadLine().ToInt32();
var march = new long[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;
}
}
BigInteger 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 | UNKNOWN | #include <bits/stdc++.h>
int main() {
long long ans;
int i, N, M = 0, A = 0, R = 0, C = 0, H = 0;
char name[11];
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 + 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;
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;
int group[5] = {};
long long int ans = 0;
string tmp;
cin >> N;
for (int i = 0; i < N; i++) {
cin >> tmp;
switch (tmp[0]) {
case 'M':
group[0]++;
break;
case 'A':
group[1]++;
break;
case 'R':
group[2]++;
break;
case 'C':
group[3]++;
break;
case 'H':
group[4]++;
break;
}
}
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 += group[i] * group[j] * group[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 i;
long long n, s = 0;
long long m = 0, a = 0, r = 0, c = 0, h = 0;
char x[100005][10];
for (i = 0; i < n; i++) {
scanf("%s", &x[i]);
}
for (i = 0; i < n; i++) {
if (x[i][0] == 'M') m++;
if (x[i][0] == 'A') a++;
if (x[i][0] == 'R') r++;
if (x[i][0] == 'C') c++;
if (x[i][0] == 'H') h++;
}
s = 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;
printf("%lld", s);
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;
string s[100010];
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> s[i];
if (s[i][0] == 'M')
m++;
else if (s[i][0] == 'A')
a++;
else if (s[i][0] == 'R')
r++;
else if (s[i][0] == 'C')
c++;
else if (s[i][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;
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 a,b;main(){std::cin>>a>>b;std::cout<<abs(a-2)*abs(b-2)<<'\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() {
long long N;
cin >> N;
vector<string> A(N);
for (long long i = 0; i < N; i++) {
cin >> A[i];
}
long long m = 0;
long long a = 0;
long long r = 0;
long long c = 0;
long long h = 0;
for (long long i = 0; i < N; i++) {
if (A[i][0] = 'M') {
m++;
} else if (A[i][0] = 'A') {
a++;
} else if (A[i][0] = 'R') {
r++;
} else if (A[i][0] = 'C') {
c++;
} else if (A[i][0] = 'H') {
h++;
}
}
long long X = 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 << X << 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;
bool sorta(pair<string, string> &a, pair<string, string> &b) {
return a.second == "dominant" || a.second == "non-existent" ||
a.second == "recessive";
}
int main() {
int t;
cin >> t;
int freq[300] = {0};
char charcter[5] = {'M', 'A', 'R', 'C', 'H'};
while (t--) {
string s;
cin >> s;
if (s[0] == 'M' || s[0] == 'A' || s[0] == 'R' || s[0] == 'C' ||
s[0] == 'H') {
freq[s[0]]++;
}
}
long long answer = 0;
for (int i = 0; i < 5; ++i) {
for (int j = i + 1; j < 5; ++j) {
for (int k = j + 1; k < 5; ++k) {
answer += (freq[charcter[i]] * freq[charcter[j]] * freq[charcter[k]]);
}
}
}
cout << answer;
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;
long long ans = 0;
vector<int> num(5, 0);
for (int i = 0; i < N; i++) {
string s;
cin >> s;
if (s[0] == 'M') num[0]++;
if (s[0] == 'A') num[1]++;
if (s[0] == 'R') num[2]++;
if (s[0] == 'C') num[3]++;
if (s[0] == 'H') num[4]++;
}
for (int i = 0; i < 5; i++) {
for (int j = i; j < 5; j++) {
for (int k = j; k < 5; k++) {
if (i == j || j == k || k == i) continue;
ans += num[i] * num[j] * num[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;
char c[5] = {'M', 'A', 'R', 'C', 'H'};
int main() {
int n;
cin >> n;
map<char, int> mp;
for (int i = 0; i < 5; i++) mp[c[i]] = 0;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
for (int i = 0; i < 5; i++) {
if (s[0] == c[i]) mp[s[0]]++;
}
}
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 += mp[c[i]] * mp[c[j]] * mp[c[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 | 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, 4, 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;
int main() {
int n, a[5] = {};
cin >> n;
while (n--) {
string s;
cin >> 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]++;
}
int cnt = 0;
for (int i = 0; i < 5; ++i) {
if (a[i] != 0) cnt++;
}
if (cnt < 3) cout << 0 << endl;
if (cnt == 3) {
long long ans = 1;
for (int i = 0; i < 5; ++i) {
if (a[i] != 0) ans *= a[i];
}
cout << ans << endl;
}
if (cnt == 4) {
long long ans = 0, as = 1;
for (int i = 0; i < 5; ++i) {
if (a[i] != 0) {
as *= a[i];
}
}
for (int i = 0; i < 5; ++i) {
if (a[i] != 0) {
ans += as / a[i];
}
}
cout << ans << endl;
}
if (cnt == 5) {
long long ans = 0;
for (int i = 0; i < 5; ++i) {
for (int j = 0; j < 5; ++j) {
if (i != j) {
long long as = 1;
for (int k = 0; k < 5; ++k) {
if (k != i && k != j) {
as *= a[k];
}
}
ans += as;
}
}
}
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.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int[] a = new int[5];
Arrays.fill(a,0);
for(int i = 0;i < N;i++){
String s = sc.next();
if(s.charAt(0) == 'M')a[0]++;
if(s.charAt(0) == 'A')a[1]++;
if(s.charAt(0) == 'R')a[2]++;
if(s.charAt(0) == 'C')a[3]++;
if(s.charAt(0) == 'H')a[4]++;
}
System.out.println(a[0]*a[1]*a[2]+a[1]*a[2]*a[3]+a[2]*a[3]*a[4]+a[3]*a[4]*a[0]+a[4]*a[0]*a[1]+a[0]*a[1]*a[3]+a[1]*a[2]*a[4]+a[2]*a[3]*a[0]+a[3]*a[4]*a[1]+a[4]*a[0]*a[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 | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
string s = "MARCH";
int c[5] = {0};
int n;
cin >> n;
for (int i = 0; i < n; i++) {
string t;
cin >> t;
for (int j = 0; j < 5; j++) {
if (t[0] == s[j]) {
c[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[i] * c[j] * c[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[n];
int cou[5];
char check[5] = {'M', 'A', 'R', 'C', 'H'};
for (int i = 0; i < 5; i++) {
cou[i] = 0;
}
for (int i = 0; i < n; i++) {
cin >> s[i];
for (int j = 0; j < 5; j++) {
if (s[i][0] == check[j]) {
cou[j]++;
}
}
}
for (int i = 0; i < 5; i++) {
cout << cou[i] << endl;
}
long long ans = 0;
for (int i = 0; i < 5; i++) {
for (int j = 0; j < 5; j++) {
for (int m = 0; m < 5; m++) {
if (j == m || i == m || i == j) {
continue;
}
ans += cou[i] * cou[j] * cou[m];
}
}
}
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() {
int n;
cin >> n;
vector<string> s(n);
for (int i = 0; i < n; i++) {
cin >> s[i];
}
long long ans = 0;
int cnt[5] = {0};
for (int i = 0; i < n; i++) {
if (s[i][0] == 'M') {
cnt[0]++;
} else if (s[i][0] == 'A') {
cnt[1]++;
} else if (s[i][0] == 'R') {
cnt[2]++;
} else if (s[i][0] == 'C') {
cnt[3]++;
} else if (s[i][0] == 'H') {
cnt[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 += 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 str;
int num[5];
for (int i = 0; i < (5); ++i) {
num[i] = 0;
}
for (int i = 0; i != N; ++i) {
cin >> str;
switch (str[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;
}
}
long long ans = 0;
for (int i = 0; i <= 2; ++i) {
for (int j = i + 1; j <= 3; ++j) {
for (int k = j + 1; k <= 4; ++k) {
ans += num[i] * num[j] * num[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 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 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 main() {
ios::sync_with_stdio(false);
cin.tie(0);
int N;
int c[5] = {0, 0, 0, 0, 0};
char tmp;
unsigned long long ans = 0;
cin >> N;
for (int i = 0; i < N; i++) {
cin >> tmp;
switch (tmp) {
case 'M':
c[0]++;
break;
case 'A':
c[1]++;
break;
case 'R':
c[2]++;
break;
case 'C':
c[3]++;
break;
case 'H':
c[4]++;
break;
}
}
for (int i = 0; i < 3; i++) {
if (c[i] == 0) continue;
for (int j = i + 1; j < 4; j++) {
if (c[j] == 0) continue;
for (int k = j + 1; j < 5; j++) {
if (c[k] == 0) continue;
ans += c[i] * c[j] * c[k];
}
}
}
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;
cin >> n;
vector<int> A(5, 0);
set<string> st;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
int set_size = st.size();
st.insert(s);
if (st.size() == set_size) continue;
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]++;
}
int 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 += A[i] * A[j] * A[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 | #include <bits/stdc++.h>
int Combination(int, int);
int main() {
int N = 0;
int M = 0, A = 0, R = 0, C = 0, H = 0;
int Z = 0;
int Result = 0;
char Input[10];
scanf("%d", &N);
for (int i = 0; i < N; i++) {
scanf("%s", Input);
if (Input[0] == 'M') M++;
if (Input[0] == 'A') A++;
if (Input[0] == 'R') R++;
if (Input[0] == 'C') C++;
if (Input[0] == 'H') H++;
}
if (M > 0)
Z++;
else
M = 1;
if (A > 0)
Z++;
else
A = 1;
if (R > 0)
Z++;
else
R = 1;
if (C > 0)
Z++;
else
C = 1;
if (H > 0)
Z++;
else
H = 1;
Result = Combination(Z, 3);
if (M > 1) Result = Result * M - Combination(Z - 1, 3) * A * R * C * H;
if (A > 1) Result = Result * A - Combination(Z - 1, 3) * M * R * C * H;
if (R > 1) Result = Result * R - Combination(Z - 1, 3) * M * A * C * H;
if (C > 1) Result = Result * C - Combination(Z - 1, 3) * M * A * R * H;
if (H > 1) Result = Result * H - Combination(Z - 1, 3) * M * A * R * C;
printf("%d", Result);
return 0;
}
int Combination(int x, int y) {
int a = 1, b = 1;
int result = 0;
for (int i = 1; i <= x; i++) {
a *= i;
}
for (int i = 1; i <= y; i++) {
b *= i;
}
for (int i = 1; i <= (x - y); i++) {
b *= i;
}
result = a / b;
return result;
}
|
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>;
const ll INF = 1LL << 60;
int main() {
string s = "MARCH";
int c[5] = {0};
int N;
cin >> N;
for (int i = 0; i < N; i++) {
string t;
cin >> t;
for (int j = 0; j < 5; j++) {
if (t[0] == s[j]) {
c[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[i] * c[j] * c[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;
vector<int> name_type(5, 0);
for (int i = 0; i < n; i++) {
string name;
cin >> name;
if (name[0] == 'M') {
name_type[0] += 1;
} else if (name[0] == 'A') {
name_type[1] += 1;
} else if (name[0] == 'R') {
name_type[2] += 1;
} else if (name[0] == 'C') {
name_type[3] += 1;
} else if (name[0] == 'H') {
name_type[4] += 1;
}
}
unsigned long long int 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 && name_type[i] != 0 &&
name_type[j] != 0 && name_type[k] != 0) {
ans += name_type[i] * name_type[j] * name_type[k];
}
}
}
}
cout << ans / 6 << 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>
const int INF = 1e9;
const int MOD = 1e9 + 7;
using namespace std;
int main() {
string key = "MARCH";
vector<int> c(5, 0);
int n;
cin >> n;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
for (int j = 0; j < 5; j++) {
if (s[0] == key[j]) c[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 += 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 | cpp | #include <bits/stdc++.h>
using namespace std;
long long m, a, r, c, h;
long long fact(int n) {
long long total = 1;
for (int i = 1; i <= n; i++) {
total = total * i;
}
return total;
}
long long func(int n, int b) { return fact(n) / fact(b); }
int main() {
int n;
cin >> n;
string s;
for (int i = 0; i < n; i++) {
cin >> s;
if (s[0] == 'M') m++;
if (s[0] == 'A') a++;
if (s[0] == 'R') r++;
if (s[0] == 'C') c++;
if (s[0] == 'H') h++;
}
long long cnt = 0;
if (m > 0) cnt++;
if (a > 0) cnt++;
if (r > 0) cnt++;
if (c > 0) cnt++;
if (h > 0) cnt++;
if (cnt < 3)
printf("0\n");
else {
long long ans = func(cnt, 3);
if (m > 0) {
ans += (m - 1) * func(cnt - 1, 2);
}
if (a > 0) {
ans += (a - 1) * func(cnt - 1, 2);
}
if (r > 0) {
ans += (r - 1) * func(cnt - 1, 2);
}
if (c > 0) {
ans += (c - 1) * func(cnt - 1, 2);
}
if (h > 0) {
ans += (h - 1) * func(cnt - 1, 2);
}
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;
char P[100001][11];
int main() {
int N, M = 0, A = 0, R = 0, C = 0, H = 0;
scanf("%d", &N);
for (int i = 0; i < N; i++) {
scanf("%s", P[i]);
if (P[i][0] == 'M') M++;
if (P[i][0] == 'A') A++;
if (P[i][0] == 'R') R++;
if (P[i][0] == 'C') C++;
if (P[i][0] == 'H') H++;
}
long long int sum = 0;
sum = (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 + A * C * H);
printf("%lld", 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 | #include <bits/stdc++.h>
int main() {
int N, i, j, k;
int CNT[5] = {0, 0, 0, 0, 0};
char S[10];
long ans = 0;
scanf("%d", &N);
for (i = 0; i < N; i++) {
scanf("%s", S);
if (S[0] == 'M') CNT[0]++;
if (S[0] == 'A') CNT[1]++;
if (S[0] == 'R') CNT[2]++;
if (S[0] == 'C') CNT[3]++;
if (S[0] == 'H') CNT[4]++;
}
for (i = 0; i < 3; i++) {
for (j = i + 1; j < 4; j++) {
for (k = j + 1; k < 5; k++) {
ans += CNT[i] * CNT[j] * CNT[k];
}
}
}
printf("%ld", 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 | java | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
long[] arr = new long[5];
int n = scan.nextInt();
for(int x = 0; x < n; x++){
String temp = scan.next();
switch(temp.charAt(0)){
case 'M':
arr[0]++;
break;
case 'A':
arr[1]++;
break;
case 'R':
arr[2]++;
break;
case 'C':
arr[3]++;
break;
case 'H':
arr[4]++;
break;
default:
break;
}
}
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 += arr[i]*arr[j]*arr[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;
int main(){
cin.tie(0),ios::sync_with_stdio(false);
int n; cin>>n;
vector<int>a(5,0);
for(int i=0;i<n;++i){
string s; cin>>s;
if(s.front()=='M')++a.at(0);
else if(s.front()=='A')++a.at(1);
else if(s.front()=='R')++a.at(2);
else if(s.front()=='C')++a.at(3);
else if(s.front()=='H')++a.at(4);
}
vector<int>v;
for(auto&i:a)if(i>0)v.emplace_back(i);
uint64_t ans=0;
for(int i=0;i<v.size();++i){
for(int j=i+1;j<v.size();++j){
for(int k=j+1;k<v.size();++k){
ans+=v.at(i)*v.at(j)*v.at(k);
}
}
}
cout<<ans<<"\n"s;
} |
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) {
int i, c0, c1, c2, N, m[5] = {0}, s;
char name[14];
const char sign[7] = "MARCH";
scanf("%d", &N);
for (i = 0; i < N; i++) {
scanf("%s", name);
for (int j = 0; j < 5; j++)
if (name[0] == sign[j]) m[j]++;
}
s = 0;
for (c0 = 0; c0 < 5; c0++)
for (c1 = c0 + 1; c1 < 5; c1++)
for (c2 = c1 + 1; c2 < 5; c2++) s += m[c0] * m[c1] * m[c2];
printf("%d\n", s);
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 | python3 | n = int(input())
march = ['M','A','R','C','H']
dic = {}
#for i in march:
# dic.setdefault(i,0)
#print(dic)
for i in range(n):
s = str(input())
if s[0] in march:
dic.setdefault(s[0],0)
dic[s[0]] += 1
print(dic)
ans = 0
if len(dic) == 3:
ans = 1
for item in dic.items():
ans *= item[1]
elif len(dic) == 4:
tmp_ans = 1
for item in dic.items():
tmp_ans *= item[1]
for item in dic.items():
ans += tmp_ans//item[1]
elif len(dic) == 5:
vl = list(dic.values())
print(vl)
for i in range(3):
for j in range(i+1,4):
for k in range(j+1,5):
ans += vl[i]*vl[j]*vl[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 <iostream>
using namespace std;
#include <algorithm>
#define REP(i,a,b) for(int i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)
#define int long long
char c[5] = {'M', 'A', 'R', 'C', 'H'};
int a[5];
int main(){
rep(i,5) a[i] = 0;
int n;
cin >> n;
rep(i,n){
char tmp[15];
cin >> tmp;
rep(j,5){
if(tmp[0] == c[j]){ a[j]++;}
}
}
int out = 0;
rep(i,5){ REP(j,i+1,5){ REP(k,j+1,5){
out += a[i] * a[j] * a[k];
}}}
cout << out <<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;
set<string> q;
set<string> a[10];
long long s[10];
int main() {
int n;
string ss;
scanf("%d", &n);
while (n--) {
cin >> ss;
if (ss[0] == 'M')
a[0].insert(ss);
else if (ss[0] == 'A')
a[1].insert(ss);
else if (ss[0] == 'R')
a[2].insert(ss);
else if (ss[0] == 'C')
a[3].insert(ss);
else if (ss[0] == 'H')
a[4].insert(ss);
}
for (int i = 0; i < 5; i++) {
s[i] = a[i].size();
}
sort(s, s + 5);
if (s[2] == 0)
cout << 0 << endl;
else if (s[1] == 0) {
printf("%lld\n", s[3] * s[2] * s[4]);
} else if (s[0] == 0) {
printf("%lld\n", s[3] * s[2] * s[4] + s[1] * s[2] * s[4] +
s[3] * s[1] * s[4] + s[1] * s[2] * s[3]);
} else {
printf("%lld\n", (s[0] + s[1]) * (s[2] + s[3] + s[4]) +
(s[1] + s[2]) * (s[3] + s[4]) + s[1] * s[3] * s[4] +
(s[3] + s[4]) * (s[0] + s[2]) + s[2] * s[3] * s[4] +
s[0] * s[3] * s[4]);
}
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> name(5);
for (int i = 0; i < N; i++) {
string tmp;
cin >> tmp;
if (tmp[0] == 'M')
name[0]++;
else if (tmp[0] == 'A')
name[1]++;
else if (tmp[0] == 'R')
name[2]++;
else if (tmp[0] == 'C')
name[3]++;
else if (tmp[0] == 'H')
name[4]++;
}
int64_t res = 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 && k != i) {
res += (name[i] * name[j] * name[k]);
}
}
}
}
res /= 6;
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;
using ll = long long;
ll INF = 1e+18;
int iINF = 1e9;
int main() {
int N;
cin >> N;
vector<int> arr(5);
for (ll i = (0); (i) < (N); i++) {
string tmp;
cin >> tmp;
char ini = tmp[0];
if (ini == 'M') arr[0] += 1;
if (ini == 'A') arr[1] += 1;
if (ini == 'R') arr[2] += 1;
if (ini == 'C') arr[3] += 1;
if (ini == 'H') arr[4] += 1;
}
int ans = 0;
for (ll i = (0); (i) < (3); i++) {
for (ll j = (i + 1); (j) < (4); j++) {
for (ll k = (j + 1); (k) < (5); k++) {
ans += arr[i] * arr[j] * arr[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 main() {
int n;
string s;
long long m, a, r, c, h;
long long d[5];
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};
cin >> n;
for (int i = 0; i < n; i++) {
cin >> s;
if (s[0] == 'M') m++;
if (s[0] == 'A') a++;
if (s[0] == 'R') r++;
if (s[0] == 'C') c++;
if (s[0] == 'H') h++;
}
d[0] = m, d[1] = a, d[2] = r, d[3] = c, d[4] = h;
long long ans = 0;
for (int i = 0; i < 10; i++) {
ans += d[P[i]] * d[Q[i]] * d[R[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>
int main() {
unsigned N;
scanf("%u\n", &N);
unsigned a = 0, b = 0, c = 0, d = 0, e = 0;
for (unsigned i = 0; i < N; ++i) {
char tmp[15];
scanf("%s", tmp);
if (tmp[0] == 'M') a++;
if (tmp[0] == 'A') b++;
if (tmp[0] == 'R') c++;
if (tmp[0] == 'C') d++;
if (tmp[0] == 'H') e++;
}
unsigned long long answer = a * (b * (c + d + e) + c * (d + e) + d * e) +
b * (c * (d + e) + d * e) + c * d * e;
printf("%llu\n", answer);
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() {
int n;
int mozi[5] = {0};
int ans = 1;
long long int zenbu = 0;
int i, j, k;
char s[11];
scanf("%d", &n);
for (i = 0; i < n; i++) {
scanf("%s", s);
switch (s[0]) {
case 'M':
mozi[0]++;
break;
case 'A':
mozi[1]++;
break;
case 'R':
mozi[2]++;
break;
case 'C':
mozi[3]++;
break;
case 'H':
mozi[4]++;
break;
}
}
for (i = 0; i < 3; i++) {
for (j = i + 1; j < 4; j++) {
for (k = j + 1; k < 5; k++) {
ans = mozi[i] * mozi[j] * mozi[k];
zenbu += ans;
}
}
}
printf("%lld", zenbu);
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;
char s[205];
int main() {
int n;
int m = 0, a = 0, r = 0, c = 0, h = 0;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%s", s);
if (s[0] == 'M') m++;
if (s[0] == 'A') a++;
if (s[0] == 'R') r++;
if (s[0] == 'C') c++;
if (s[0] == 'H') h++;
}
int res = m * a * (r + c + h) + m * r * (c + h) + m * c * h +
a * r * (c + h) + a * c * h + r * c * h;
printf("%d", res);
}
|
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
from itertools import combinations
march = list("MARCH")
march = dict(zip(march, march))
N = int(input())
S = [march.get(input()[0], -1) for _ in range(N)]
cnt = Counter(S)
ans = 0
for comb in combinations("MARCH", 3):
maxv = 0
for c in comb:
if cnt[c] == 0:
break
maxv = max(maxv, cnt[c])
else:
ans += maxv
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 x, y, z, c;
int d[5];
for(int i=0;i<5;i++){
d[i]=0;
}
char a[100001][100];
char b[5][2]={"M",
"A",
"R",
"C",
"H"};
cin >> x;
for (y = 0; y < x; y++) {
cin >> a[y];
for(z=0;z<5;z++){
c = strcmp(a[y][0], b[z][0]);
if (c == 0) {
d[z]=d[z]+1;
}
}
}
cout << d[0]*d[1]*d[2]*d[3]*d[4] << 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 | n = gets.chomp.split.to_i
cnt_hash = {M: 0, A: 0, R: 0, C: 0, H: 0}
persons = n.times map
gets.chomp
end
grouped_p = persons.group_by{|p| p[0]}
.map{|k,v| [k.to_sym, v.size]}.to_h
cnt_hash.merge! grouped_p
ans = "MARCH".chars.combination(3).inject(0) do |sum, attr|
sum + cnt_hash[attr[0].to_sym] * cnt_hash[attr[1].to_sym] * cnt_hash[attr[2].to_sym]
end
|
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[n];
vector<int> c(5);
for (int i = 0; i < n; i++) {
cin >> s[i];
switch (s[i][0]) {
case 'M':
c[0]++;
break;
case 'A':
c[1]++;
break;
case 'R':
c[2]++;
break;
case 'C':
c[3]++;
break;
case 'H':
c[4]++;
break;
}
}
long ans = 0;
ans += c[0] * c[1] * c[2];
ans += c[0] * c[1] * c[3];
ans += c[0] * c[1] * c[4];
ans += c[0] * c[2] * c[3];
ans += c[0] * c[2] * c[4];
ans += c[0] * c[3] * c[4];
ans += c[1] * c[2] * c[3];
ans += c[1] * c[2] * c[4];
ans += c[1] * c[3] * c[4];
ans += c[2] * c[3] * c[4];
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 n, a[100], ans;
string s[10000];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) cin >> s[i];
for (int i = 1; i <= n; i++) {
if (s[i][0] == 'M') ++a[1];
if (s[i][0] == 'A') ++a[2];
if (s[i][0] == 'R') ++a[3];
if (s[i][0] == 'C') ++a[4];
if (s[i][0] == 'H') ++a[5];
}
ans += a[1] * a[2] * a[3];
ans += a[1] * a[2] * a[4];
ans += a[1] * a[2] * a[5];
ans += a[1] * a[3] * a[4];
ans += a[1] * a[3] * a[5];
ans += a[1] * a[4] * a[5];
ans += a[2] * a[3] * a[4];
ans += a[2] * a[3] * a[5];
ans += a[3] * a[4] * a[5];
printf("%d\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 | java | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int[] S = new int[5];
for (int i = 0; i < N; i++) {
String name = sc.next();
if (name.charAt(0) == 'M') S[0]++;
if (name.charAt(0) == 'A') S[1]++;
if (name.charAt(0) == 'R') S[2]++;
if (name.charAt(0) == 'C') S[3]++;
if (name.charAt(0) == 'H') S[4]++;
}
long comb = 0;
for (int i = 0; i < 5; i++) {
for (int j = i+1; j < 5; j++) {
for (int k = j+1; k < 5; k++) {
comb += S[i] * S[j] * S[k];
}
}
}
System.out.println(comb);
}
}
|
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 int N, X = 0;
string S, T = "MARCH";
cin >> N;
vector<int> A(5, 0);
for (int i = 0; i < N; i++) {
cin >> S;
for (int j = 0; j < 5; j++)
if (S[0] == T[j]) A.at(j)++;
}
for (int i = 2; i < 5; i++)
for (int j = 1; j < i; j++)
for (int k = 0; k < j; k++) X += A.at(i) * A.at(j) * A.at(k);
cout << X << 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 | #icnlude <bits/stdc++.h>
using namespace std;
int main() {
int m,a,r,c,h,n;
cin >> n;
m = 0;
a = 0;
r = 0;
c = 0;
h = 0;
vector<string> names(n);
for(int i=0;i<n;i++){
cin >> names.at(i);
if(names.at(i).at(0) == 'M'){
m++;
}
else if(names.at(i).at(0) == 'A'){
a++;
}
else if(names.at(i).at(0) == 'R'){
r++;
}
else if(names.at(i).at(0) == 'C'){
c++;
}
else if(names.at(i).at(0) == 'H'){
h++;
}
}
long x;
x = 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 << x << 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;
long long ll;
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]++;
}
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 += 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;
template <class A>
void pr(A a) {
cout << a;
cout << '\n';
}
template <class A, class B>
void pr(A a, B b) {
cout << a << ' ';
pr(b);
}
template <class A, class B, class C>
void pr(A a, B b, C c) {
cout << a << ' ';
pr(b, c);
}
template <class A, class B, class C, class D>
void pr(A a, B b, C c, D d) {
cout << a << ' ';
pr(b, c, d);
}
template <class A>
void PR(A a, long long n) {
for (long long i = (long long)(0); i < (long long)(n); i++) {
if (i) cout << ' ';
cout << a[i];
}
cout << '\n';
}
long long check(long long n, long long m, long long x, long long y) {
return x >= 0 && x < n && y >= 0 && y < m;
}
const long long MAX = 1e9 + 7, MAXL = 1LL << 61, dx[4] = {-1, 0, 1, 0},
dy[4] = {0, 1, 0, -1};
void Main() {
map<char, long long> m;
int n;
cin >> n;
for (long long i = (long long)(0); i < (long long)(n); i++) {
string s;
cin >> s;
m[s[0]]++;
}
vector<long long> v;
for (__typeof((m).begin()) it = (m).begin(); it != (m).end(); it++)
v.push_back(it->second);
long long ans = 0;
for (long long i = (long long)(0); i < (long long)(v.size()); i++) {
for (long long j = (long long)(i + 1); j < (long long)(v.size()); j++) {
for (long long k = (long long)(j + 1); k < (long long)(v.size()); k++)
ans += v[i] * v[j] * v[k];
}
}
pr(ans);
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
Main();
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>
void swap(long long *a, long long *b) {
long long c;
c = *b;
*b = *a;
*a = c;
}
long long max2(long long a, long long b) { return a >= b ? a : b; }
long long min2(long long a, long long b) { return a >= b ? b : a; }
long long ABS(long long a) { return a >= 0 ? a : (-a); }
typedef struct {
long long aa;
long long bb;
} frequent;
int compare(const void *a, const void *b) {
return *(long long *)a - *(long long *)b;
}
int main(void) {
long long n, i, ans = 0;
scanf("%lld", &n);
char s[10], ini[n];
for (i = 0; i < n; i++) {
scanf("%s", s);
ini[i] = s[0];
}
long long letter[] = {0, 0, 0, 0, 0};
for (i = 0; i < n; i++) {
if (ini[i] == 'M') letter[0]++;
if (ini[i] == 'A') letter[1]++;
if (ini[i] == 'R') letter[2]++;
if (ini[i] == 'C') letter[3]++;
if (ini[i] == 'H') letter[4]++;
}
for (i = 0; i < n; i++) {
for (long long j = i + 1; j < n; j++) {
for (long long k = j + 1; k < n; k++) {
ans += letter[i] * letter[j] * letter[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 ll = long long;
using namespace std;
int main() {
int N;
cin >> N;
vector<int> M(5, 0);
string S;
for (int i = 0; i < (N); i++) {
cin >> S;
if (S[0] == 'M') {
M[0]++;
} else if (S[0] == 'A') {
M[1]++;
} else if (S[0] == 'R') {
M[2]++;
} else if (S[0] == 'C') {
M[3]++;
} else if (S[0] == 'H') {
M[4]++;
}
}
ll sum = 0;
for (int i = 0; i < (3); i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
sum += M[i] * M[j] * M[k];
}
}
}
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 <stdio.h>
#include <algorithm>
using namespace std;
struct comb{
int a, b, c;
ordenar(){
int d=a, e=b, f=c;
if(d > e){
int aux = d;
d = e;
e = aux;
}
if(f < e){
if(f > d){
int aux = e;
e = f;
f = aux;
} else {
int aux = f;
f = e;
e = d;
d = aux;
}
}
a = d;
b = e;
c = f;
}
};
int n, abc[6], cont = 0, com[4], values[6], idx = 0;
char word[20];
comb tot[200];
void f(int p, int v){
if(p == 2){
com[p] = v;
tot[idx].a = com[0];
tot[idx].b = com[1];
tot[idx].c = com[2];
tot[idx].ordenar();
idx+=1;
return;
}
com[p] = v;
values[v] = 1;
for(int i=1; i<=cont; i++){
if(values[i] == 0){
f(p+1, i);
}
}
values[v] = 0;
}
bool compare(comb uno, comb dos){
if(uno.a == dos.a){
if(uno.b == dos.b){
return uno.c < dos.c;
} else {
return uno.b < dos.b;
}
} else {
return uno.a < dos.a;
}
}
int main(){
scanf("%d", &n);
for(int i=0; i<n; i++){
scanf("%s", word);
if(word[0] == 'M')
abc[0] += 1;
else if(word[0] == 'A')
abc[1] += 1;
else if(word[0] == 'R')
abc[2] += 1;
else if(word[0] == 'C')
abc[3] += 1;
else if(word[0] == 'H')
abc[4] += 1;
}
int x=1, y=1, m=1, id=0;
for(int i=0; i<5; i++){
if(abc[i] > 0){
cont += 1;
m*=abc[i];
abc[id] = abc[i];
id+=1;
}
}
for(int i=1; i<=cont; i++)
x = x*i;
for(int i=1; i<=(cont - 3); i++)
y = y*i;
x = (x/(y*6));
for(int i=1; i<=cont; i++){
f(0, i);
}
sort(tot, tot+idx, compare);
//values[0] = tot[0].a; values[1] = tot[0].b; values[2] = tot[0].c;
int k=0;
for(int i=0; i<idx; i++){
if(tot[i].a == values[0] && tot[i].b== values[1] && tot[i].c == values[2]){
} else {
values[0] = tot[i].a; values[1] = tot[i].b; values[2] = tot[i].c;
int z=abc[values[0]-1]*abc[values[1]-1]*abc[values[2]-1];
k += z;
//printf("%d\n", z);
//printf("%d %d %d\n", tot[i].a, tot[i].b, tot[i].c);
}
}
printf("%d\n", 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 | n = int(input())
s = []
for i in range(n):
s.append(input())
tmp_nums = {"M":0, "A":0, "R":0, "C":0, "H":0}
for i in range(n):
tmp_nums[s[i][0]] += 1
nums=[]
for i in tmp_nums.values():
nums.append(i)
p = [0,0,0,0,0,0,1,1,1,2]
q = [1,1,1,2,2,3,2,2,3,3]
r = [2,3,4,3,4,4,3,4,4,4]
sum = 0
for i in range(10):
sum += nums[p[i]]*nums[q[i]]*nums[r[i]]
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;
int mp[1000];
long long res;
int n;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
char ch[1000];
scanf("%s", ch);
mp[ch[0]]++;
}
int P[6] = {0, 'M', 'A', 'R', 'C', 'H'};
for (int i = 1; i <= 3; i++)
for (int j = i + 1; j <= 4; j++)
for (int k = j + 1; k <= 5; k++) res += mp[P[i]] * mp[P[j]] * mp[P[k]];
printf("%lld\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;
const int N = 100010;
const double eps = 1e-6;
int n;
vector<string> s[5];
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
string t;
cin >> t;
if (t[0] == 'A')
s[0].push_back(t);
else if (t[0] == 'C')
s[1].push_back(t);
else if (t[0] == 'H')
s[2].push_back(t);
else if (t[0] == 'M')
s[3].push_back(t);
else if (t[0] == 'R')
s[4].push_back(t);
}
int cnt = 0;
for (int i = 0; i < 5; i++) cnt += s[i].size();
if (cnt < 3) {
puts("0");
return 0;
}
long long res = 0;
for (int i = 0; i < 5; i++)
for (int j = i + 1; j < 5; j++)
for (int k = j + 1; k < 5; k++) {
int a = s[i].size(), b = s[j].size(), c = s[k].size();
res += a * b * c;
}
cout << res << 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<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<vector>
using namespace std;
using ll = long long;
#define MM = 1000000000;
#define mod = MM + 7;
#define INF (ll)1e18
#define pi acos(-1.0)
#define MAX 100005
#define NIL -1
int main(){
ll n; cin >> n;
string s;
map<char, ll> mp;
ll cnt = 0; //C用
ll ans = 1; //ans
char G = {'M', 'A', 'R', 'C', 'H'};
for(ll i = 0; i < n; i++){
cin >> s;
if(s[0]=='M'||s[0]=='A'||s[0]=='R'||s[0]=='C'||s[0]=='H'){
mp[s[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 *= mp[G[i]]*mp[G[j]]*mp[G[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 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 << i << "," << j << "," << k << endl;
}
}
}
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>
const int PRIME = 1000000007;
const int INF = 2147483647;
const double PI = acos(-1);
const double EPS = 0.000000001;
using namespace std;
int main() {
int i, n, cnts[5], cnt = 0;
for (i = 0; i < 5; i++) cnts[i] = 0;
cin >> n;
for (i = 0; i < n; i++) {
string tmp;
cin >> tmp;
if (tmp[0] == 'M') cnts[0]++;
if (tmp[0] == 'A') cnts[1]++;
if (tmp[0] == 'R') cnts[2]++;
if (tmp[0] == 'C') cnts[3]++;
if (tmp[0] == 'H') cnts[4]++;
}
for (int a = 0; a < 3; a++) {
for (int b = a + 1; b < 4; b++) {
for (int c = b + 1; c < 5; c++) {
cnt += cnts[a] * cnts[b] * cnts[c];
}
}
}
cout << cnt << 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 | python3 | n = int(input())
s = [list(input()) for i in range(n)]
ans = 0
for i in range(n):
for j in range(i+1,n):
for k in range(j+1,n):
p = s[i][0]
q = s[j][0]
r = s[k][0]
if (p=='M'or'A'or'C'or'R'or'H') and (q=='M'or'A'or'C'or'R'or'H') and (r=='M'or'A'or'C'or'R'or'H') and (p!=q) and (q!=r) and (r!=p):
ans += 1
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;
cin >> n;
vector<int> a(5, 0);
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s.at(0) == 'M') a.at(0)++;
if (s.at(0) == 'A') a.at(1)++;
if (s.at(0) == 'R') a.at(2)++;
if (s.at(0) == 'C') a.at(3)++;
if (s.at(0) == 'H') a.at(4)++;
}
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 += a.at(i) * a.at(j) * a.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 | cpp | #include <bits/stdc++.h>
using namespace std;
string march("MARCH");
int main() {
int N;
cin >> N;
vector<string> S(N);
copy_n(istream_iterator<string>(cin), N, S.begin());
unordered_map<char, int> m;
for (auto& item : S) {
if (string::npos != march.find_first_of(item[0], 0)) {
if (m.count(item[0]))
++(m[item[0]]);
else
m[item[0]] = 1;
}
}
int64_t result = 0;
for (size_t i = 0; i < march.size() - 2; ++i) {
for (size_t j = i + 1; j < march.size() - 1; ++j) {
for (size_t k = j + 1; k < march.size(); ++k) {
result += m[march[i]] * m[march[j]] * m[march[k]];
}
}
}
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 | cpp | #include <bits/stdc++.h>
using namespace std;
int main(int argc, const char* argv[]) {
string nameHead = "MARCH";
int str[5] = {0};
int N;
cin >> N;
for (int i = 0; i < N; i++) {
string inp;
cin >> inp;
for (int j = 0; j < 5; j++) {
if (nameHead[j] == inp[0]) str[j]++;
}
}
int sum = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
sum += str[i] * str[j] * str[k];
}
}
}
cout << sum << 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;
int64_t cnt = 0;
cin >> n;
map<char, int64_t> mp;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
char a = s.at(0);
mp[a]++;
}
int64_t b = mp['M'];
int64_t c = mp['A'];
int64_t d = mp['R'];
int64_t e = mp['C'];
int64_t f = mp['H'];
cnt += b * (c * d + c * e + c * f + d * e + d * f + e * f) +
c * (d * e + d * f * +e * f) + d * e * f;
cout << cnt << 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 vin = vector<int>;
using P = pair<int, int>;
const int inf = 1e9 + 7;
const ll INF = 1e18;
int main() {
ll n;
cin >> n;
vector<string> s(n);
for (int i = 0; i < (n); i++) cin >> s[i];
ll ans = 0;
map<char, int> mp;
for (int i = 0; i < (n); i++) {
mp[s[i][0]]++;
}
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 Graph = vector<vector<int>>;
const long long MOD = 1e9 + 7;
int main() {
int n;
cin >> n;
int m = 0, a = 0, r = 0, c = 0, h = 0;
for (int i = 0; i < (int)(n); i++) {
string s;
cin >> s;
if (s[0] == 'M') m++;
if (s[0] == 'A') a++;
if (s[0] == 'R') r++;
if (s[0] == 'C') c++;
if (s[0] == 'H') h++;
}
long long ans = 0;
ans += m * a * r;
ans += m * a * c;
ans += m * a * h;
ans += m * r * c;
ans += m * r * h;
ans += m * c * h;
ans += a * r * c;
ans += a * r * h;
ans += a * c * h;
ans += 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>
int main() {
int n;
char name[11];
int avoid = scanf("%d", &n);
int i;
int count[5] = {};
for (i = 0; i < n; i++) {
avoid = scanf("%s", name);
switch (name[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]++;
}
}
long long int ans;
ans = count[0] * count[1] * count[2] + count[0] * count[1] * count[3] +
count[0] * count[1] * count[4] + count[0] * count[2] * count[3] +
count[0] * count[2] * count[4] + count[0] * count[3] * count[4] +
count[1] * count[2] * count[3] + count[1] * count[2] * count[4] +
count[1] * count[3] * count[4] + count[2] * count[3] * count[4];
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 | python3 | import 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 | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using Pi = pair<int, int>;
using Pl = pair<ll, ll>;
int main() {
int N;
cin >> N;
vector<string> S(N);
vector<int> sum(5);
for (int i = (int)(0); i < (int)(N); i++) {
cin >> S[i];
if (S[i][0] == 'M') sum[0]++;
if (S[i][0] == 'A') sum[1]++;
if (S[i][0] == 'R') sum[2]++;
if (S[i][0] == 'C') sum[3]++;
if (S[i][0] == 'H') sum[4]++;
}
ll ans = 0;
for (int i = (int)(0); i < (int)(3); i++) {
for (int j = (int)(i + 1); j < (int)(4); j++) {
for (int k = (int)(j + 1); k < (int)(5); k++) {
ans += sum[i] * sum[j] * sum[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 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;
}
}
int 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", 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;
template <class T>
T ceil(T a, T b) {
return (a + b - 1) / b;
}
template <class T>
T round(T a, T b) {
return (a + b / 2) / b;
}
template <class T>
T gcd(T a, T b) {
return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
return a * b / gcd(a, b);
}
template <class T>
bool amax(T& a, const T& b) {
if (a < b) {
a = b;
return true;
}
return false;
}
template <class T>
bool amin(T& a, const T& b) {
if (b < a) {
a = b;
return true;
}
return false;
}
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
int N;
cin >> N;
map<char, int> M = {{'M', 0}, {'A', 1}, {'R', 2}, {'C', 3}, {'H', 4}};
int cnt[5] = {};
for (int i = (int)(0); i < (int)(N); i += (int)(1)) {
string S;
cin >> S;
if (S[0] == 'M' || S[0] == 'A' || S[0] == 'R' || S[0] == 'C' ||
S[0] == 'H') {
cnt[M[S[0]]]++;
}
}
int res = 0;
for (int i = (int)(0); i < (int)(3); i += (int)(1)) {
for (int j = (int)(i + 1); j < (int)(5); j += (int)(1)) {
for (int k = (int)(j + 1); k < (int)(5); k += (int)(1)) {
res += cnt[i] * cnt[j] * cnt[k];
}
}
}
cout << res << 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 pi = pair<int, int>;
using vi = vector<int>;
const ll linf = 1LL << 62;
const int inf = 99999999;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const ll atcoder = 1e9 + 7;
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
std::fill((T *)array, (T *)(array + N), val);
}
ll gcd(ll a, ll b) {
if (a < b) swap(a, b);
if (a % b == 0)
return b;
else
gcd(b, a % b);
}
ll lcm(ll a, ll b) {
if (a < b) swap(a, b);
return (a / gcd(a, b)) * b;
}
int main() {
int n;
string t = "MARCH";
int a[5] = {0};
cin >> n;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
for (int j = 0; j < t.length(); j++)
if (s[0] == t[j]) a[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 += a[i] * a[j] * a[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;
vector<int> v(5, 0);
long long ans = 0;
for (int i = 0; i < (int)(n); i++) {
cin >> s;
if (s[0] == 'M') {
v[0]++;
}
if (s[0] == 'A') {
v[1]++;
}
if (s[0] == 'R') {
v[2]++;
}
if (s[0] == 'C') {
v[3]++;
}
if (s[0] == 'H') {
v[4]++;
}
}
sort(v.begin(), v.end());
vector<int> per = {0, 0, 1, 1, 1};
do {
int tmp = 1;
for (int i = 0; i < 5; i++) {
if (per[i] != 0) tmp *= v[i] * per[i];
}
ans += tmp;
} while (next_permutation(per.begin(), per.end()));
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 | python3 | n = int(input())
l = [0]*7
ans = 0
for i in range(n):
s = input()
if s[0] == "M":
l[0] += 1
elif s[0] == "A":
l[1] += 1
elif s[0] == "R":
l[2] += 1
elif s[0] == "C":
l[3] += 1
elif s[0] == "H":
l[4] += 1
for i in range(2):
for j in range(i+1,i+3):
for k in range(j+1,j+3):
ans += l[i] * l[j] * l[k]
# print(i,j,k,ans)
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 <iostream>
#include <stack>
#include <algorithm>
#include <vector>
#include <math.h>
#include <stdio.h>
#include <functional>
#include <string>
#include <cstdlib>
#include <numeric>
#include <cstdbool>
#include <map>
#include<set>
#include<queue>
typedef long long ll;
#define rep(i,n) for(ll i=0;i<n;i++)
using namespace std;
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; }
const long long INF = 1000000000000000;
const ll inf = -1e18;
typedef pair<int, int> P;
ll ma = 1000000000 + 7;
ll h, w, n,k,m; string s,t;
char maze[60][60];
int dis[101][101]; int graph[52][52]; bool vis[52];
int dx[4] = { 1, 0, -1, 0 };
int dy[4] = { 0, 1, 0, -1 };
ll gcd(ll x, ll y) {
if (x % y == 0) return y;
return gcd(y, x % y);
}
ll lcm(ll a,ll b) {
ll g = gcd(a, b);
return a / g * b;
}
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() {
cin >> n; vector<int> mo(5); ll res = 0;
rep(i, n) {
string a;
cin >> a;
if (a[0] == 'M')mo[0]++;
else if (a[0] == 'A')mo[1]++;
else if (a[0] == 'R')mo[2]++;
else if (a[0] == 'C')mo[3]++;
else if (a[0] == 'H')mo[4]++;
}
rep(i, 10) {
res += mo[P[i]] * mo[Q[i]] * mo[R[i]];
}
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 | UNKNOWN | #include <bits/stdc++.h>
int main() {
long a[5] = {0, 0, 0, 0, 0};
int N;
scanf("%d", &N);
int k;
for (k = 0; k < N; k++) {
char s[10];
scanf("%s", 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]++;
}
int O = 0;
int u, n, ko;
for (u = 0; u <= 2; u++) {
for (n = u + 1; n <= 3; n++) {
for (ko = n + 1; ko <= 4; ko++) {
O += a[u] * a[n] * a[ko];
}
}
}
printf("%d", O);
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;
scanf("%d", &n);
char s[n][11];
int mem[5];
for (int i = 0; i < 5; i++) {
mem[i] = 0;
}
for (int i = 0; i < n; i++) {
scanf("%s", &s[i]);
if (s[i][0] == 'M') mem[0]++;
if (s[i][0] == 'A') mem[1]++;
if (s[i][0] == 'R') mem[2]++;
if (s[i][0] == 'C') mem[3]++;
if (s[i][0] == 'H') mem[4]++;
}
long long int 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 += mem[i] * mem[j] * mem[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 | UNKNOWN | num = gets.to_i
initial = {"M": 0,"A": 0,"R": 0,"C": 0,"H": 0}
person = []
while num > 0 do
num -= 1
n = gets.to_s[0]
if n == "M"||"A"||"R"||"C"||"H"
initial[n.to_sym] += 1
end
end
initial = initial.select {|k, v| v > 0 }
#{"M": 2,"A": 2,"R": 1,"C": 1}
initial2 = initial.select {|k, v| v > 1 }
#{"M": 2,"A": 2}
#MAR MAC ARC
count = 0
"MARCH".split("").combination(3).map do |a,b,c|
count = count + (sum[a] * sum[b] * sum[c])
end
puts count
|
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 cnt[5] = {0};
long long int ans = 0;
int main() {
int N;
char s[11] = {0};
scanf("%d", &N);
for (int i = 0; i < N; i++) {
scanf("%s", s);
if (s[0] == 'M')
cnt[0]++;
else if (s[0] == 'A')
cnt[1]++;
else if (s[0] == 'R')
cnt[2]++;
else if (s[0] == 'C')
cnt[3]++;
else if (s[0] == 'H')
cnt[4]++;
}
for (int i = 0; i <= 2; i++) {
for (int j = i + 1; j <= 3; j++) {
for (int k = j + 1; k <= 4; 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 | UNKNOWN | #[allow(unused_macros)]
macro_rules! input {
(source = $s:expr, $($r:tt)*) => {
let mut iter = $s.split_whitespace();
let mut next = || { iter.next().unwrap() };
input_inner!{next, $($r)*}
};
($($r:tt)*) => {
let stdin = std::io::stdin();
let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
let mut next = move || -> String{
bytes
.by_ref()
.map(|r|r.unwrap() as char)
.skip_while(|c|c.is_whitespace())
.take_while(|c|!c.is_whitespace())
.collect()
};
input_inner!{next, $($r)*}
};
}
#[allow(unused_macros)]
macro_rules! input_inner {
($next:expr) => {};
($next:expr, ) => {};
($next:expr, $var:ident : $t:tt $($r:tt)*) => {
let $var = read_value!($next, $t);
input_inner!{$next $($r)*}
};
}
#[allow(unused_macros)]
macro_rules! read_value {
($next:expr, ( $($t:tt),* )) => {
( $(read_value!($next, $t)),* )
};
($next:expr, [ $t:tt ; $len:expr ]) => {
(0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
};
($next:expr, chars) => {
read_value!($next, String).chars().collect::<Vec<char>>()
};
($next:expr, usize1) => {
read_value!($next, usize) - 1
};
($next:expr, $t:ty) => {
$next().parse::<$t>().expect("Parse error")
};
}
#[allow(unused_imports)]
use std::cmp::{min, max};
fn main() {
input!{
n: usize,
ss: [chars; n],
}
let mut c = vec![0; 5];
for i in 0..n {
match ss[i][0] {
'M' => c[0] += 1,
'A' => c[1] += 1,
'R' => c[2] += 1,
'C' => c[3] += 1,
'H' => c[4] += 1,
_ => continue,
}
}
let uc = c.iter().filter(|&&v| v >= 1).count() as u64;
if uc < 3 {
println!("{}", 0);
return
}
let mut ans = 0;
for i in 0..5 {
for j in i+1..5 {
for k in j+1..5 {
ans += c[i] * c[j] * c[k];
}
}
}
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;
int main() {
int a, c, d, e, f = 0, ans = 0, n, m;
int b;
int M = 0, A = 0, R = 0, C = 0, H = 0;
int num[100000];
unsigned long long x, y;
string s;
cin >> n;
b = 0;
for (a = 0; a < n; a++) {
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++;
}
}
a = 0;
a += M * A * R;
a += M * A * C;
a += M * A * H;
a += M * R * C;
a += M * R * H;
a += M * C * H;
a += A * R * C;
a += A * R * H;
a += A * C * H;
a += R * C * H;
cout << a << 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 solve(int x[]) {
int ans = 0;
ans += x[0] * x[1] * x[2];
ans += x[0] * x[1] * x[3];
ans += x[0] * x[1] * x[4];
ans += x[0] * x[2] * x[3];
ans += x[0] * x[2] * x[4];
ans += x[0] * x[3] * x[4];
ans += x[1] * x[2] * x[3];
ans += x[1] * x[2] * x[4];
ans += x[1] * x[3] * x[4];
ans += x[2] * x[3] * x[4];
return ans;
}
int main() {
int n;
int march[5];
for (int i = 0; i < 5; i++) {
march[i] = 0;
}
string s;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> s;
if (s[0] == 'M') {
march[0] += 1;
} else if (s[0] == 'A') {
march[1] += 1;
} else if (s[0] == 'R') {
march[2] += 1;
} else if (s[0] == 'C') {
march[3] += 1;
} else if (s[0] == 'H') {
march[4] += 1;
} else {
continue;
}
}
int ans = solve(march);
cout << 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.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
final int N = sc.nextInt();
int ary[] = {0, 0, 0, 0, 0}; //M, A, R, C, H
for(int i=0; i<N; i++) {
char c = sc.next().charAt(0);
switch(c) {
case 'M': ary[0]++; break;
case 'A': ary[1]++; break;
case 'R': ary[2]++; break;
case 'C': ary[3]++; break;
case 'H': ary[4]++;
}
}
sc.close();
long x = 0L;
for(int i=0; i<=2; i++) {
for(int j=i+1; j<=3; j++) {
for(int k=j+1; k<=4; k++) {
x += ary[i] * ary[j] * ary[k];
}
}
}
System.out.println(x);
}
} |
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 d, m, a, r, c, h, b[10], e;
string s[100010], f[100010];
cin >> d;
for (int i = 0; i < a; i++) {
cin >> s[i];
}
for (int i = 0; i < 11; i++) {
b[i] = 0;
}
m = 1, a = 2, r = 3, c = 4, h = 5;
for (int i = 0; i < a; i++) {
if (s[i][0] == 'M') b[m]++;
if (s[i][0] == 'A') b[a]++;
if (s[i][0] == 'R') b[r]++;
if (s[i][0] == 'C') b[c]++;
if (s[i][0] == 'H') b[h]++;
}
e = 0;
for (int i = 0; i < 11; i++) {
for (int j = i + 1; j < 11; j++) {
for (int k = j + 1; k < 11; k++) {
e += b[i] * b[j] * b[k];
}
}
}
cout << e << 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 ll MOD = 1e9 + 7;
const ll INF = 1e18;
using Graph = vector<vector<int>>;
int main() {
int n;
cin >> n;
string s;
map<int, int> memo;
for (int i = 0; i < n; i++) {
cin >> s;
if (s[0] == 'M')
memo[s[0] - 'A']++;
else if (s[0] == 'A')
memo[s[0] - 'A']++;
else if (s[0] == 'R')
memo[s[0] - 'A']++;
else if (s[0] == 'C')
memo[s[0] - 'A']++;
else if (s[0] == 'H')
memo[s[0] - 'A']++;
}
ll ans = 0;
for (int i = 0; i < 24; i++) {
for (int j = i + 1; j < 25; j++) {
for (int k = j + 1; k < 26; k++) {
ans += memo[i] * memo[j] * memo[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;
long long n, q, k, x, y, z, dist[111111];
vector<pair<long long, long long> > g[111111];
void dfs(long long i, long long fa) {
for (long long j = 0; j < g[i].size(); j++) {
long long to = g[i][j].first, val = g[i][j].second;
if (to == fa) continue;
dist[to] = dist[i] + val;
dfs(to, i);
}
}
int main() {
cin >> n;
for (long long i = 1; i < n; i++) {
cin >> x >> y >> z;
g[x].push_back(make_pair(y, z));
g[y].push_back(make_pair(x, z));
}
cin >> q >> k;
dfs(k, 0);
while (q--) {
cin >> x >> y;
cout << dist[x] + dist[y];
}
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 mod = 1e9 + 7;
int main() {
int n;
cin >> n;
string s = "MARCH";
int c[5] = {0};
for (int i = 0; i < (n); i++) {
string tmp;
cin >> tmp;
for (int j = 0; j < (n); j++) {
if (tmp[0] == s[j]) {
c[j]++;
}
}
}
long long anser = 0;
for (int i = 0; i < (5); i++) {
for (int j = i + 1; j < (5); j++) {
for (int k = j + 1; k < (5); k++) {
anser += c[i] * c[j] * c[k];
}
}
}
cout << anser << 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;
char name[5] = {'M', 'A', 'R', 'C', 'H'};
int possible[5] = {0, 0, 0, 0, 0};
int count = 0;
int func(int x, int n) {
if (n == 1) return x;
return x * func(x - 1, n - 1);
}
int f(int x) {
if (x == 0) return 1;
return x * f(x - 1);
}
int main() {
int n;
cin >> n;
vector<string> s(n);
vector<int> memo(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] == name[j]) {
possible[j]++;
}
}
}
int people = 0;
for (int i = 0; i < 5; i++) {
if (possible[i] > 0) {
people++;
}
}
if (people < 3) {
cout << 0 << endl;
return 0;
}
long long int res = 0;
for (int i = 0; i < 5; i++) {
for (int j = 0; j < 5; j++) {
if (j <= i) continue;
for (int k = 0; k < 5; k++) {
if (k <= j) continue;
res += possible[i] * possible[j] * possible[k];
}
}
}
cout << res << 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 | n = gets.to_i
s = []
n.times do
s << gets.chop
end
s.sort!
alpha = {}
"MARCH".chars do |c|
alpha[c] = 0
end
s.map do |x|
alpha[x[0]] += 1
end
count = []
"MARCH".chars do |c|
count << alpha[c]
end
ans = 0
count.combination(3).map do |c|
ans += c.inject(:*)
end
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;
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;
}
const long long INF = 1000000000000000;
const long long inf = -1e18;
long long ma = 1000000000 + 7;
long long h, w, n, k, m;
string s, t;
char maze[60][60];
int dis[101][101];
int graph[52][52];
bool vis[52];
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
long long gcd(long long x, long long y) {
if (x % y == 0) return y;
return gcd(y, x % y);
}
long long lcm(long long a, long long b) {
long long g = gcd(a, b);
return a / g * b;
}
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() {
cin >> n;
vector<int> mo(5);
long long res = 0;
for (long long i = 0; i < n; i++) {
string a;
cin >> a;
if (a[0] == 'M')
mo[0]++;
else if (a[0] == 'A')
mo[1]++;
else if (a[0] == 'R')
mo[2]++;
else if (a[0] == 'C')
mo[3]++;
else if (a[0] == 'H')
mo[4]++;
}
for (long long i = 0; i < 10; i++) {
res += mo[p[i]] * mo[Q[i]] * mo[R[i]];
}
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 <iostream>
#include <string>
#include <vector>
using namespace std;
int main(){
long long int N;
string s;
cin>>N;
vector<ll> v(5);
for(ll i=0;i<N;i++){
cin>>s;
if(s[0]=='M') v[0]++;
else if(s[0]=='A') v[1]++;
else if(s[0]=='R') v[2]++;
else if(s[0]=='C') v[3]++;
else if(s[0]=='H') v[4]++;
}
cout<<v[0]*(v[1]*(v[2]+v[3]+v[4])+v[2]*(v[3]+v[4])+v[3]*v[4])+v[1]*(v[2]*(v[3]+v[4])+v[3]*v[4])+v[2]*v[3]*v[4]<<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 | python3 | import itertools
total = int(input())
lis1 = [input() for x in range(total)]
lis2 = []
for name in lis1:
if "M" == name[0] or "A" == name[0] or "R" == name[0] or "C" == name[0] or "H" == name[0]:
lis2.append(name)
count = 0
for a in itertools.combinations(lis2,3):
if a[0][0] != a[1][0] and a[1][0] != a[2][0] and a[0][0] != a[2][0]:
count += 1
print(count)
|
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>
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define ll long long
#define pb(a) push_back(a)
#define INF 999999999
using namespace std;
typedef pair<int, int> PA;
typedef pair<ll, ll> LAP;
typedef pair<int, PA> PPA;
typedef pair<ll, LAP> LPPA;
int dy[]={0, 0, 1, -1, 0};
int dx[]={1, -1, 0, 0, 0};
ll int P[10] = {0, 0, 0, 0, 0, 0, 1, 1, 1, 2};
ll int Q[10] = {1, 1, 1, 2, 2, 3, 2, 2, 3, 3};
ll int R[10] = {2, 3, 4, 3, 4, 4, 3, 4, 4, 4};
int main()
{
ll int N;
string name;
cin >> N;
ll int M = 0, A = 0, R = 0, C = 0, H = 0;
REP(i,N)
{
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;
}
ll int temp[5];
temp[0] = M;
temp[1] = A;
temp[2] = R;
temp[3] = C;
temp[4] = H;
ll int ans = 0;
REP(i,10) ans += temp[P[i]] * temp[Q[i]] * temp[R[i]];
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() {
int n;
cin >> n;
long long m = 0, a = 0, r = 0, c = 0, h = 0;
for (int i = 0; i < n; i++) {
string str;
cin >> str;
char ch = str[0];
if (ch == 'm') m++;
if (ch == 'a') a++;
if (ch == 'r') r++;
if (ch == 'c') c++;
if (ch == 'h') h++;
}
long long sum = 0;
sum += m * a * r;
sum += m * a * c;
sum += m * a * h;
sum += m * r * c;
sum += m * r * h;
sum += m * c * h;
sum += a * r * c;
sum += a * r * h;
sum += a * c * h;
sum += r * c * h;
cout << sum << 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;
using System.Collections.Generic;
using System.Numerics;
namespace AtCoder
{
class Program
{
static void Main(string[] args)
{
MainStream();
}
static void MainStream()
{
Scan sc = new Scan();
write wr = new write();
int[] march = new int[5];
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[] R = { 2, 3, 4, 3, 4, 4, 3, 4, 4, 4 };
int n = sc.intarr[0];
for (int i = 0; i < n; i++)
{
string name = sc.str;
char init = name[0];
if (init == 'M')
{
march[0]++;
}
if (init == 'A')
{
march[1]++;
}
if (init == 'R')
{
march[2]++;
}
if (init == 'C')
{
march[3]++;
}
if (init == 'H')
{
march[4]++;
}
}
BigInteger result = 0;
for(int i=0; i<10; i++)
{
result += march[P[i]] * march[Q[i]] * march[R[i]];
}
Console.WriteLine(result);
}
}
class Scan
{
public string str => Console.ReadLine();
public string[] strarr => str.Split(' ');
public long[] longarr => strarr.Select(long.Parse).ToArray();
public int[] intarr => strarr.Select(int.Parse).ToArray();
public char[] chararr => str.ToArray();
}
class write
{
public void wri<Type>(Type x)
{
Console.WriteLine(x);
}
}
class Method
{
static public int max(params int[] nums)
{
int maxx = nums[0];
for (int i = 1; i < nums.Length; i++)
{
maxx = Math.Max(maxx, nums[i]);
}
return maxx;
}
}
}
|
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;
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 (int i = 0; i < (3); ++i) {
for (int j = i + 1; j < 5 - 1; 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 | cpp | #include<bits/stdc++.h>
#define ll long long
#define endl '\n'
const ll mod = 1000000007;
const double PI = 3.141592653589793238463;
using namespace std;
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
// long long m=0,r=0,a=0,c=0,h=0;
int x;
int A[5]= {0};
cin>>x;
string s;
set<string>se;
while(x--)
{
cin>>s;
if(s[0]=='M')
{
if(count(se.begin(),se.end(),s)==0)
A[0]++;
se.insert(s);
}
if(s[0]=='R')
{
if(count(se.begin(),se.end(),s)==0){
A[1]++;
se.insert(s);
}
}
if(s[0]=='A')
{
if(count(se.begin(),se.end(),s)==0)}{
A[2]++;
se.insert(s);
}
}
if(s[0]=='C')
{
if(count(se.begin(),se.end(),s)==0){
A[3]++;
se.insert(s);
}
}
if(s[0]=='H')
{
if(count(se.begin(),se.end(),s)==0){
A[4]++;
se.insert(s);
}
}
}
int sum=0;
for(int i=0; i<3; i++)
{
for(int j=i+1; j<4; j++)
{
for(int k=j+1; k<5; k++)
{
if(A[i]!=0&&A[j]!=0&&A[k]!=0)
sum+=(A[i]*A[j]*A[k]);
}
}
}
cout<<sum;
return 0;
}
|
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