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.
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).
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
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);
}
}