hdu2058The sum problem

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]
看了网上的基本都是根据长度算出。先确定一段和sum＝m的最长的长度，要最长则起点从1开始，根据等差求和，len*(len+1)/2=m;放缩法则len^2<2*m,所以len=pow(2.0*m,0.5);现在根据长度确定起点L，根据等差求和公式，((len+L-1)+L)*len/2=m; =>  L=(2*m/len+1-len)/2; 判断时反代上式是否成立，是则输出[L,L+len-1。
正确代码：
#include<stdio.h>#include<math.h>int main(){    int n,m,l;    while(scanf("%d%d",&n,&m)>0&&n+m!=0)    {        for(int len=(int)pow(2.0*m,0.5);len>=1;len--)        {            l=(2*m/len-len+1)/2;            if(len*(len+2*l-1)/2==m)            printf("[%d,%d]\n",l,l+len-1);        }        printf("\n");    }}



超时代码：
#include<stdio.h>int main(){    int n,m,l,r,mid;    while(scanf("%d%d",&n,&m)>0&&n+m!=0)    {        m=m<n?m:n;        for(int i=1;i<m&&2*i+1<=m;i++)        {            l=i;r=m;            while(l<=r)            {                mid=(l+r)/2;                if(2*m==mid*(mid+1)+i-i*i)                {                    printf("[%d,%d]\n",i,mid);break;                }                if(2*m<mid*(mid+1)+i-i*i) r=mid-1;                if(2*m>mid*(mid+1)+i-i*i) l=mid+1;            }        }        if(m<=n)        printf("[%d,%d]\n",m,m);        printf("\n");    }}