【题解】- hdu The sum problem【技巧】

The sum problem

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 10
50 30
0 0

Sample Output

[1,4]
[10,10]

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

批注

不可使用暴力枚举，会超时，应从等差数列求和的公式下手，先找到最大的长度

AC代码

#include <stdio.h>
#include <math.h>
int main(void)
{
    long long n, m, i, h, k;//假设所求区间为[k,h];
    while(scanf("%lld%lld",&n,&m) != EOF){
        if(n == 0 && m == 0){
            break;
        }
        for(i = sqrt(2 * n); i > 0; i --){//那么假设i为区间的长度即从k到h的长度
            k = ( 2 * m / i + 1 - i ) / 2;//根据等差数列前n项和公式（首项+末项）乘于a（区间长度）再除于2就等于m了(即：(((j+i)*a)/2=m))因为这里等差为1所以i=j+a-1 ==> ((j+a-1+j)*a)/2=m ==> j=(2*m/a+1-a)/2所以这里直接暴力枚举就OK了
            h = k + i - 1;
            if(k == 0) continue;
            if(m == i * k + i * (i - 1) / 2){
                printf("[%lld,%lld]\n", k, h);
            }
        }
        printf("\n");
    }
    return 0;
}