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第一轮 F
Multiple of 17Time Limit:1000MS Memory Limit:0KB 64bit IO Format:%lld & %lluSubmit StatusDescriptionDownload as PDF Multiple of 17 Theorem: If you drop the last digit d of an integer n (n$ \ge$10), subtract 5d from the remaining integer, then the difference is a multiple of 17 if and only if n is a multiple of 17.For example, 34 is a multiple of 17, because 3-20=-17 is a multiple of 17; 201 is not a multiple of 17, because 20-5=15 is not a multiple of 17.Given a positive integer n, your task is to determine whether it is a multiple of 17.Input There will be at most 10 test cases, each containing a single line with an integer n ( 1$ \le$n$ \le$10100). The input terminates with n = 0, which should not be processed.Output For each case, print 1 if the corresponding integer is a multiple of 17, print 0 otherwise.Sample Input 34201209876541317171717171717171717171717171717171717171717171717180Sample Output 1010Problemsetter: Rujia Liu, Special Thanks: Yiming Li/************************************************************************* > File Name: f.cpp > Author:yuan > Mail: > Created Time: 2014年11月09日 星期日 13时04分13秒 ************************************************************************/ #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> using namespace std; int mat[105]; char str[105]; int ans; int main() { while(1){ scanf("%s",str); if(str[0]=='0') break; int l=strlen(str); for(int i=0;i<l;i++) { mat[i]=str[i]-'0'; } ans=0; for(int i=0;i<l-1;i++) { ans=ans*10+mat[i]; ans=ans%17; } ans=(ans-mat[l-1]*5)%17; if(ans==0) printf("1\n"); else printf("0\n"); } return 0; }
第一轮 F
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