HDOJ 题目4602 Partition（找规律，快速幂）

Partition

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

Total Submission(s): 2635 Accepted Submission(s): 1052



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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).

[align=left]Output[/align]
Output the required answer modulo 109+7 for each test case, one per line.

[align=left]Sample Input[/align]

2
4 2
5 5

[align=left]Sample Output[/align]

5
1

[align=left]Source[/align]
2013 Multi-University Training Contest 1

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1 2 3 4 5

1 1 2 5 12 28

2 1 2 5 12

3 1 2 5

4 1 2

5 1

规律：2^(m-3)*(m-2). m=n-k+1

ac代码

#include<stdio.h>
#include<string.h>
#define mod 1000000007
__int64 qpow(__int64 a,__int64 b)
{
__int64 ans=1;
while(b)
{
if(b&1)
ans=(ans*a)%mod;
a=(a*a)%mod;
b/=2;
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
__int64 n,k;
scanf("%I64d%I64d",&n,&k);
if(k>n)
{
printf("0\n");
continue;
}
if(n-k+1<=2)
{
printf("%d\n",n-k+1);
}
else
{
printf("%I64d\n",(qpow(2,n-k-2)*(n-k+3))%mod);
}
}
}