大数取模

2019-04-13 21:08发布

这里记一下两种方法 1、字符串 #include #include #include using namespace std; typedef long long ll; int main(){ char num[100005]; int i, j, x; while(~scanf("%s %d", num,&x)){ ll ans = 0; for(i = 0; i < strlen(num); ++i){ ans = ans*10 + (num[i]-'0'); ans %= x; } printf("%lld ",ans); getchar(); } return 0; }
1、JAVA(本人不擅长JAVA。。。选自大牛博客) import java.io.*; import java.util.*; import java.math.BigInteger; public class Main { public static void main(String args[]) { BigInteger big1 = new BigInteger("0"); BigInteger big2 = new BigInteger("0"); Scanner cin = new Scanner (System.in); while(cin.hasNext()) { big1 = cin.nextBigInteger(); big2 = cin.nextBigInteger(); big1 = big1.remainder(big2); System.out.println(big1); } } }JAVA代码原地址:http://m.blog.csdn.net/ysc504/article/details/8818279


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