HDU 4602 partition (Probability Method)

Partition Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 2472 accepted submission (s): 978


Problem descriptiondefine 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.
 
Inputthe 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 ).
 
Outputoutput the required answer modulo 109 + 7 for each test case, one per line.
Sample Input

24 25 5

 
Sample output

51

For 1 <= k <n, we can arrange the N points into a column and take out K consecutive points. In this case, the K ends are disconnected consecutively;

1. If the K points contain the endpoint (we only consider an endpoint), there are still (n-k-1) intervals,

Each interval has two states: disconnected and closed, so there are 2 ^ (n-k-1), and multiplied by 2;

2, if the K points do not contain the endpoint, the continuous K points have (n-k-1) method, there are (n-k-2) interval,

So there are 2 ^ (n-k-2) * (n-k-1 );

Total 2? 2 ^ (n-k? 1) + 2 ^ (n-k? 2 )? (N-k? 1) = (n-k + 3) * 2 ^ (n-k? 2 ).

#include"stdio.h"#include"string.h"#include"queue"#include"vector"#include"algorithm"using namespace std;#define LL __int64const int mod=1000000007;int main(){    int T,i,n,k;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&k);        if(n<k)            printf("0\n");        else if(n==k)            printf("1\n");        else        {            LL d=n-k,s1,t;            if(d==1) printf("2\n");            else            {                s1=d+3;                d-=2;                LL aa=2,tmp=1;                while(d)                {                    if(d&1)                        tmp*=aa;                    d/=2;                    aa=(aa*aa)%mod;                    tmp%=mod;                }                printf("%I64d\n",(tmp*s1)%mod);            }        }    }    return 0;}

HDU 4602 partition (Probability Method)