J - The sum problem
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J - The sum problem
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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.
InputInput contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0.
OutputFor 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 10
50 30
0 0
Sample Output
[1,4]
[10,10] [4,8]
[6,9]
[9,11]
[30,30] 數學思維題目，，看了一下大佬的博文。。感覺 還復雜啊！！
#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<math.h>
using namespace std;
/*
思路:
區間項數為i
起始位置 a,終點為b則b=a+i-1;
這一段區間的和為sum=(a+b)*i/2=(a+a+i-1)*i/2=m;
a最小為1，則（1+1+i-1）*i/2<=m即(i+1)*i<=2m , i有一定小于sqrt(2m)
即 0<i<sqrt(2m);判斷條件就是 (a+a+i-1)*i==2*m;
*/
int main()
{
int n,m,a,b;
while(cin>>n>>m)
{
if(n== && m==)
break;
for(int i=sqrt(*m);i>=;i--)//最小為1項，就是[m,m]
{
a=m/i-(i-)/;//等差公式 即 m=a*i+i*(i-1)/2-----m/i=a+(i-1)/2---a=m/i-(i-1)/2;
if((a+a+i-)*i==*m)
{
printf("[%d,%d]\n",a,a+i-);
}
}
printf("\n");
}
return ;
}
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