Calculate the Sum

最新推荐文章于 2023-12-04 11:52:52 发布

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最新推荐文章于 2023-12-04 11:52:52 发布

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Calculate the Sum

Time Limit: 1000MS	Memory Limit: 10000KB
Submissions: 72	Accepted: 17

Description

As you all know, MOD is a mathematical operatio. Giving you two numbers n,m（0 < m,n <= 10^10001），Your task is to calculate the sum of every digit of m MOD every digit of n. We can guarantee that there is no zero in digits of n.

Input

The first line, a integer T, representing T test cases blew.(T<=10).

In each case, the first line is the integer m, the second line is the integer n.

Output

Print the sum of every digit of m MOD every digit of n.

Sample Input

1
13
21

Sample Output

Hint

Source

SWUST 7th ACM/ICPC Programming Contest - 2011

数据很大，10^10001次方，如果直接2重循环去遍历，肯定超过1ms，所以必须巧妙处理，考虑到每位的数字是0到9之间的，可以开一个数组，记录一下前一个数据中0到9这些数字分别有几个，这样就可以降低复杂度了。

#include<stdio.h>
#include<string.h>
char a[10001],b[10001];
int main()
{
    int t,i,j,sum,lena,lenb,c[10];
scanf("%d",&t);
while(t--)
{
  sum=0;
  memset(c,0,sizeof(c));
   scanf("%s%s",a,b);
      lena=strlen(a);
   lenb=strlen(b);
   for(i=0;i<lena;i++)
    ++c[a[i]-'0'];
   for(i=0;i<lenb;i++)
    for(j=0;j<10;j++)
     sum+=j%(b[i]-'0')*c[j];
   printf("%d\n",sum);
}
}