lightoj--1214--Large Division(大数取余)
Large Division
Submit Status Description Given two integers,a and b,you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c. Input Input starts with an integer T (≤ 525),denoting the number of test cases. Each case starts with a line containing two integers a (-10200 ≤ a ≤ 10200) and b (|b| > 0,b fits into a 32 bit signed integer). Numbers will not contain leading zeroes. Output For each case,print the case number first. Then print 'divisible' if a is divisible by b. Otherwise print 'not divisible'. Sample Input 6 101 101 0 67 -101 101 7678123668327637674887634 101 11010000000000000000 256 -202202202202000202202202 -101 Sample Output Case 1: divisible Case 2: divisible Case 3: divisible Case 4: not divisible Case 5: divisible Case 6: divisible Source
Problem Setter: Jane Alam Jan
水题一枚 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; char s[1001]; int main() { int t,k=1; scanf("%d",&t); while(t--) { memset(s,' ',sizeof(s)); scanf("%s",s); long long mod,temp; scanf("%lld",&mod); temp=0; for(int i=0;i<strlen(s);i++) { if(s[i]=='-') continue; temp=temp*10+(s[i]-'0'); temp%=mod; } if(temp%mod) printf("Case %d: not divisiblen",k++); else printf("Case %d: divisiblen",k++); } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |