题目链接:http://acm.hust.edu.cn/vjudge/contest/125004#problem/D
密码:acm
Description
If an integer is not divisible by 2 or 5, some multiple of that number in decimal notation is a sequence of only a digit. Now you are given the number and the only allowable digit, you should report the number of digits of such multiple.
For example you have to find a multiple of 3 which contains only 1's. Then the result is 3 because is 111 (3-digit) divisible by 3. Similarly if you are finding some multiple of 7 which contains only 3's then, the result is 6, because 333333 is divisible by 7.
Input
Input starts with an integer T (≤ 300), denoting the number of test cases.
Each case will contain two integers n (0 < n ≤ 106 and n will not be divisible by 2 or 5) and the allowable digit (1 ≤ digit ≤ 9).
Output
For each case, print the case number and the number of digits of such multiple. If several solutions are there; report the minimum one.
Sample Input Sample Output
Case :
Case :
Case :
分析:
第一次代码,超了应该,反正看输不出结果:
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std; int main()
{
int n,i,o=,a,b;
scanf("%d", &n); for(i=;i<=n;i++)
{
scanf("%d%d", &a,&b);
int d=b,ans=; while(d%a)
{
d=d*+b;
ans++;
} printf("Case %d: %d\n", o++,ans);
}
return ;
}
修改后:
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std; int main()
{
int n,i,o=,a,b;
scanf("%d", &n); for(i=;i<=n;i++)
{
scanf("%d%d", &a,&b);
int d=b,ans=,r; r=b%a;
while(r)
{
d=r*+b;
ans++;
r=d%a;
} printf("Case %d: %d\n", o++,ans);
}
return ;
}