題目：

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.

題意：給出一個數N，找有多少個數，這些數的大小小于2^N，且這些數符合2^k-1的形式。

思路：把100以內的這種數和對應的N輸出出來，會發現個數與N是成三倍的關系的。

#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; } View Code

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轉載于:https://www.cnblogs.com/sykline/p/9747475.html

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