hdoj #2058
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.
Simple Input
20 10
50 30
0 0
Simple Output
[1,4]
[10,10]
[4,8]
[6,9]
[9,11]
[30,30]
方法：等差求和公式的运用
Sn = n * a0 + (n * (n - 1) / 2) * d; n为项数，d为公差，a0为首项；
1+2+3+....+(2m)^1/2 > m 可由等差求和公式证明；所以最多就(2m)^1/2项；
代码如下：
#include <iostream>
#include <cmath>
using namespace std;
int main(){
int n, m;
while(cin>>n>>m && (n || m)){
for(int i = sqrt(2 * m);i >= 1;i--){
int a = (m - (i*(i-1))/2) / i;
if((a * 2 + i - 1) * i == 2 * m)
printf("[%d,%d]
", a, a+i-1);
}
printf("
");
}
return 0;
}