1023.Have Fun with Numbers (20)

/**
* pat-al-1023
* 2017-01-26
* C version
* Author: fengLian_s
*/
#include<stdio.h>
#include<string.h>
int main()
{
//test:
// int a = 123456789;
// a *= 2;
// printf("a * 2 = %d\n", a);
// a *= 2;
// printf("a * 2 = %d\n", a);
//test end
freopen("in.txt", "r", stdin);
char num[22];//用long long不行，后三个测试点过不了，应该是位数超限了
scanf("%s", num);
char newNum[22];
newNum[strlen(num)] = '\0';//坑：不用long long，换成字符串处理之后，如果不加这句，后三个点能过，前面的又过不了了。不懂为什么后三个点能过？
int tmpNum, c = 0;//c是进位
int hs1[10] = {0}, hs2[10] = {0};
for(int i = strlen(num)-1;i >= 0;i--)
{
tmpNum = (num[i] - '0') * 2 + c;
c = tmpNum / 10;//更新进位的值
newNum[i] = tmpNum % 10 + '0';
hs1[num[i] - '0']++;
hs2[newNum[i] - '0']++;
}
if(c > 0)//对最高位进位的特殊处理
hs2[c]++;
int flagFail = 0;
for(int i = 0;i < 10;i++)
{
if(hs1[i] != hs2[i])
{
flagFail = 1;
break;
}
}
if(flagFail)
printf("No\n");
else
printf("Yes\n");
if(c > 0)
printf("%d", c);
printf("%s\n", newNum);
return 0;
}