HDU 4602 Partition

Partition

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1388 Accepted Submission(s): 573



Problem Description

Define f(n) as the number of ways to perform n in format of the sum of some positive integers. For instance, when n=4, we have

4=1+1+1+1

4=1+1+2

4=1+2+1

4=2+1+1

4=1+3

4=2+2

4=3+1

4=4

totally 8 ways. Actually, we will have f(n)=2(n-1) after observations.

Given a pair of integers n and k, your task is to figure out how many times that the integer k occurs in such 2(n-1) ways. In the example above, number 1 occurs for 12 times, while number 4 only occurs once.

Input

The first line contains a single integer T(1≤T≤10000), indicating the number of test cases.

Each test case contains two integers n and k(1≤n,k≤109).

Output

Output the required answer modulo 109+7 for each test case, one per line.

Sample Input

2
4 2
5 5

Sample Output

5
1

Source

2013 Multi-University Training Contest 1

Recommend

liuyiding

题意： 略：

思路： 推到出 2 ^ (m - 3) * (m + 2) m = n - k + 1;
用快速矩阵幂

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

const int V = 100000 + 50;
const int MaxN = 100000 + 5;
const int mod = 1000000000 + 7;
int T, n, k;
__int64 pow_mod(__int64 b) {
__int64 a = 2;
__int64 res = 1;
while(b) {
if(b & 1)
res = res * a % mod;
b >>= 1;
a = a * a % mod;
}
return res % mod;
}
int main() {
int i, j;
scanf("%d", &T);
while(T--) {
scanf("%d%d", &n, &k);
int m = n - k + 1;
if(k > n)
printf("0\n");
else if(m <= 2)
printf("%d\n", m);
else if(m == 3)
printf("5\n");
else
printf("%I64d\n", (pow_mod(m - 3) * (m + 2) % mod) % mod);
}
}