Codeforces--50B--Choosing Symbol Pairs

题目描述：
There is a given string S consisting of N symbols. Your task is to find the number of ordered pairs of integers i and j such that
1 ≤ i, j ≤ N
S[i] = S[j], that is the i-th symbol of string S is equal to the j-th.
输入描述:
The single input line contains S, consisting of lowercase Latin letters and digits. It is guaranteed that string S in not empty and its length does not exceed 105.
输出描述:
Print a single number which represents the number of pairs i and j with the needed property. Pairs (x, y) and (y, x) should be considered different, i.e. the ordered pairs count.
输入：
great10
aaaaaaaaaa
输出：
7
100
题意：
计算字符串中Si = Sj的次数
题解：
对于每一种字符，它的个数为t，则它的贡献为t * t
代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<map>
using namespace std;const int maxn = 100000 + 5;
char s[maxn];
typedef long long ll;int main(){while(scanf("%s",s)!=EOF){map<char,int> mp;int l = strlen(s);for(int i = 0; i < l; i ++){mp[s[i]] ++;}ll sum = 0;map<char, int>::iterator it;for(it = mp.begin(); it != mp.end(); it++){int t = it->second;sum += 1ll * t * t;}printf("%I64d\n",sum);}return 0;
}

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