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You are given two 10-based integers b and x, and you are required to determine the following proposition is true or false:
For arbitrary b-based positive integer y=c1c2⋯cn¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ (ci is the i-th dight from left of y), define f(y)=∑i=1nci, if f(f(⋯f(y)⋯))∞ can be divided by x, then y can be divided by x, otherwise y can’t be divided by x
.
Input
The first line contains a 10-based integer t (1≤t≤105) — the number of test cases.
For each test case, there is a single line containing two 10-based integers b and x (2≤b,x≤1018)
.
Output
For each test case, if the proposition is true, print “T”, otherwise print “F” (without quotes).
Sample Input
1
10 3
Sample Output
T
题解
来自多校联赛
理解上面的,代码就特别简单了
代码:
#include<stdio.h>
int main()
{int t;scanf("%d",&t);while(t--){long long b,x;scanf("%lld %lld",&b,&x);if(b%x==1)printf("T\n");elseprintf("F\n");}return 0;
}
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