HDU2058 找规律+递推

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The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21732 Accepted Submission(s): 6380

Problem Description

Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.

Input

Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0.

Output

For each test case, print all the possible sub-sequence that its sum is M.The format is show in the sample below.print a blank line after each test case.

Sample Input

20 1050 300 0

Sample Output

[1,4][10,10][4,8][6,9][9,11][30,30]

等差数列求和的问题，不过这里数据量过大所以要剪枝

我一开始想的是2*i肯定大于m，但是还是超了，看了人家的优化方案是这样的

1+2+。。。+（2*m)^1/2>m

#include <iostream>#include <math.h>using namespace std;int main(){    __int64 m ,n;    while(cin>>n>>m,n||m)    {        __int64 a,k;        for(k=(int)sqrt(2*m);k>0;k--)//因为1+2+。。。+（2*m)^1/2>m;所一和为m的元素个数应该小于等于（2*m）^1/2;        {            a=m/k-(k-1)/2;//假设从a开始一直加到a+k-1,一共k个元素和为m,可以计算出a;            if((2*a-1+k)*k==2*m) //（2*m)^1/2可能不是整数，判断是有必要的；                cout<<"["<<a<<","<<k+a-1<<"]"<<endl;        }cout<<endl;    }return 0;}

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