Multiple of 17
Time Limit:1000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
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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
34
201
2098765413
1717171717171717171717171717171717171717171717171718
0
Sample Output
1
0
1
0
Problemsetter: 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
");
else printf("0
");
}
return 0;
}