Number Theory Problem（The 2016 ACM-ICPC Asia China-Final Contest 找规律）

题目：

Mr. Panda is one of the top specialists on number theory all over the world. Now Mr. Panda is investigating the property of the powers of 2. Since 7 is the lucky number of Mr. Panda, he is always interested in the number that is a multiple of 7. However, we know that there is no power of 2 that is a multiple of 7, but a power of 2 subtracted by one can be a multiple of 7. Mr. Panda wants to know how many positive integers less than 2ⁿ in the form of 2^k-1 (k is a positive integer) that is divisible by 7. N is a positive interger given by Mr Panda.

Input：

The first line of the input gives the number of test cases, T. T test cases follow. Each test case contains only one positive interger N.

output:

For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the answer.

题意：给出一个数N，找有多少个数，这些数的大小小于2^N，且这些数符合2^k-1的形式。

思路：把100以内的这种数和对应的N输出出来，会发现个数与N是成三倍的关系的。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
using namespace std;
vector<long double> v;

int read()
{
    int res = 0;
    char op = getchar();
    if(op>='0' && op<='9')
    {
        res = op-'0';
        op = getchar();
    }
    while(op>='0'&&op<='9')
    {
        res = res*10+op-'0';
        op = getchar();
    }
    return res;
}
int main()
{
    int T,n,cnt=1;
    T = read();
    while(T--)
    {
        int n;
        n = read();
        printf("Case #%d: %d\n",cnt++,n/3);
    }
    return 0;
}

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