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1097: 求余

Description

小學題目，給定除數和被除數，求余數

Input

多組測試數據,每組測試數據包含兩個整數n,k(1<=n<=10^2000,1<=k<=1000)

Output

對於每組測試數據,輸出n%k

Sample Input

12 4 12 5

Sample Output

0 2 思路：同余定理，（a+b) % m = ((a % m) + (b % m)) % m 。 代碼如下： #include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<time.h>
using namespace std;
#define FORA(i,x,y) for(int i = x; i < y; i++)
#define FORB(i,x,y) for(int i = x; i <= y; i++)
#define FORC(i,y,x) for(int i = y; i > x; i--)
#define maxn 100000
#define INF 1000000000
#define LL long long
const int mod = 1000000;
char a[maxn];
int b[maxn];
int main(){
int n,m;
while(~scanf("%s %d",a,&m)){
int t = strlen(a);
int ans = 0;
FORA(i,0,t){
ans = (ans * 10 + a[i] - ‘0‘) % m;
}
printf("%d\n",ans);
}
return 0;
}

zcmu 1097