标签:rem als cout code 题目 include log php space
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1212
题意:给出两个数a, b,求a%b;
思路:(c+d)%e=c%e+d%e,(c*d)%e=(c%e*d%e)%e;
代码:
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 #define ll long long
5 #define MAXN 1000+10
6 using namespace std;
7
8 int main(void){
9 ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
10 int b;
11 char a[MAXN];
12 while(~scanf("%s%d", a, &b)){
13 int ans=0, cnt=1;
14 for(int i=strlen(a)-1; i>=0; i--){
15 int gg=(a[i]-‘0‘)*cnt;
16 ans=(ans+gg)%b;
17 cnt=cnt*10%b;
18 }
19 printf("%d\n", ans);
20 }
21 return 0;
22 }
据说java大数很厉害。。然后get了一下。。果然威力无穷。。。
代码:
1 import java.math.BigDecimal;
2 import java.util.Scanner;
3
4 public class Main{
5 public static void main(String args[]){
6 Scanner scan = new Scanner(System.in);
7 while(scan.hasNext()){
8 BigDecimal a = scan.nextBigDecimal();
9 int b = scan.nextInt();
10 System.out.println(a.remainder(new BigDecimal(b)));
11 }
12 }
13 }
标签:rem als cout code 题目 include log php space
原文地址:http://www.cnblogs.com/geloutingyu/p/5989883.html