hd1212

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.
To make the problem easier, I promise that B will be smaller than 100000.
Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.

 Input

The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.

 

Output

For each test case, you have to ouput the result of A mod B.

 

Sample Input

2 312 7152455856554521 3250

 

Sample Output

251521
 
题目大意就是说给一很长的数字，需要用数组来存的，再给一小于100000的数，求字符串表示的数对这个数的余数。
不会，然后看的人家用的同余定理结合秦九韶公式来做的，然后才写出现在的代码
 
//4)同余定理a+b≡x+m （mod d）//其中 a≡x (mod d)，b≡m(mod d)//6)a*b≡x*m (mod d )//其中a≡x (mod d),b≡m (mod d)//秦九韶算法f(x)=a(n)x^n+a(n-1)x^(n-1)+....=x(a(n)x^(n-1)+a(n-1)x^(n-2)+... #include<stdio.h>#include<string.h>char a[1001];int aa[1001];int main(){int b,lena,i,y;while(scanf("%s %d",a,&b)==2){lena=strlen(a);for(i=0;i<lena;++i)aa[i]=a[i]-'0';                  y=0;for(i=1;i<lena;++i){aa[i]+=(aa[i-1]*10)%b;}y=aa[lena-1]%b;printf("%d\n",y);}return 0;}