Calculate the Sum
| Time Limit: 1000MS | Memory Limit: 10000KB |
| Submissions: 72 | Accepted: 17 |
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.
Sample Input
1 13 21
Sample Output
2
數據很大，10^10001次方，如果直接2重循環去遍歷，肯定超過1ms，所以必須巧妙處理，考慮到每位的數字是0到9之間的，可以開一個數組，記錄一下前一個數據中0到9這些數字分別有幾個，這樣就可以降低複雜度了。
#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);
}
}