PAT 1023 Have Fun with Numbers（简单计数+大整数）

题目

https://www.patest.cn/contests/pat-a-practise/1023

题意：输入一串不超过20位的数字，将该数字串乘以两倍后，判断新数串是否为原数串的一个排列。

解题思路

统计旧字符串中数字0-9的出现次数和新字符串中数字0-9的出现次数，若两者完全相同，则新数串是原数串的一个排列。

AC代码

#include <iostream>#include <cstring>using namespace std;int main(){    char str[25];    cin >> str;    int s[25], index = 0;    int old_cnt[10] = {0};    for (int i = strlen(str)-1; i >= 0; --i)    {        s[index++] = str[i]-'0'; //store digits inversely        old_cnt[str[i]-'0']++; //count each digit    }    //double    int res[25] = {0}, new_cnt[10] = {0}, tmp;    for (int i = 0; i < index; ++i)    {        tmp = s[i] << 1;        res[i] += tmp%10;        res[i+1] += tmp/10;    }    if (res[index] != 0)        new_cnt[res[index]]++;    for (int i = 0; i <= index; ++i)    {        if (i != index)            new_cnt[res[i]]++;        else if (res[index] != 0)            new_cnt[res[index]]++;    }    //check    bool flag = true;    for (int i = 0; i <= 9; ++i)    {        if (old_cnt[i] != new_cnt[i])        {            flag = false;            break;        }    }    if (flag) cout << "Yes" << endl;    else cout << "No" << endl;    for (int i = index; i>=0; --i)    {        if (i == index && res[index] == 0) //leading digit is zero            continue;        cout << res[i];    }    cout << endl;    return 0;}