PAT 1023. Have Fun with Numbers (20)（字符串转换，20位数的加法）

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1023. Have Fun with Numbers (20)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line “Yes” if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or “No” if not. Then in the next line, print the doubled number.

Sample Input:
1234567899
Sample Output:
Yes
2469135798

解题思路

• 1.判断和是否由原来的数子组成。

AC代码

#include<iostream>#include<string>#include<vector>#include<sstream>using namespace std;int keep[10];int main(int argc, char *argv[]){    string a,b;    cin >> a;    //keep置0    fill(keep,keep+10,0);    b = "";    int next = 0;    for (int i = a.length()-1; i >= 0; i--) {        int tem = a[i] - '0';        //keep记录每个数字的个数        keep[tem]++;        tem = 2 * tem + next;        if (tem > 9) {            tem = tem - 10;            next = 1;        }else {            next = 0;        }        char c = '0' + tem;        b = c + b;    }    //最高位相加大于10的情况    if (next > 0) {        char c = '0' + next;        b = c + b;    }    bool flag =true;    //判断b中的数字是不是都是a中的    for (int i = 0; i <b.length() ; ++i) {        int tem = b[i] - '0';        if (keep[tem] <= 0) {            cout << "No" << endl;            flag = false;            break;        }else {            keep[tem] --;        }    }    if (flag) {        cout << "Yes" << endl;    }    cout << b << endl;    return 0;}