大数相乘
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 就是判断前边的数字是否可以被后边的整除,,要按照字符串的格式处理 #include<iostream> #include<string> using namespace std; int main(){ int t,k=0; cin>>t; while(t--){ k++; string a; int b; cin>>a>>b; long long ans=0,i=0; if(a[0]==‘-‘) i++; for(;i<a.size();i++){ ans=(ans*10+a[i]-‘0‘)%b; } printf("Case %d: ",k); if(ans==0)puts("divisible"); else puts("not divisible"); } 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! |