Choosing Symbol Pairs

寻找符号对的数量

最新推荐文章于 2024-12-17 14:17:09 发布

原创最新推荐文章于 2024-12-17 14:17:09 发布 · 670 阅读

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该问题要求找到字符串S中满足特定条件的有序符号对的数量。给定一个包含小写字母和数字的字符串S，任务是计算有多少对(i, j)，使得S[i]等于S[j]。输入是一个不超过105个字符的非空字符串。输出是满足条件的符号对数量，注意(i, j)和(j, i)被视为不同的对。示例输入和输出分别展示了如何计算。" 77822062,5762741,使用JavaScript创建人物行走动画,"['javascript', '动画']

Description
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.

Input
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.

Output
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.

Sample Input
Input
great10
Output
7
Input
aaaaaaaaaa
Output
100
题目很水，就是让你配对，题意也很好理解，注意int不行，会越界,我错了两次就是没有进行好强制转换，看一下我的代码吧

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <algorithm>
using namespace std;
int main()
{
    char x[100005];
    int a[50];
    while(~scanf("%s", x))
    {
        memset(a, 0, sizeof(a));
        int len = strlen(x);
        for(int i = 0; i < len; i++)
        {
            if(islower(x[i])) a[x[i] - 'a' + 10]++;
            else a[x[i] - '0']++;
        }
        long long sum = 0;
        for(int i = 0; i < 36; i++)
        {
            sum += (a[i] * 1LL *a[i]);
        }
        printf("%lld\n", sum);
    }
    return 0;
}