http://acm.hdu.edu.cn/showproblem.php?pid=2058
20 10 50 30 0 0
[1,4] [10,10] [4,8] [6,9] [9,11] [30,30]
找一个子序列的和等于M,那么这个子序列可以看成a+1, a+2, ... , a+d 这时,d就为这个序列的长度,起始数字就是a+1,而这个序列的和即M=a*d + (1 + d) * d /2;得出d*d<2 * m,从而可以枚举d,计算出a
#include<iostream> #include<cmath> using namespace std; int main(){ int n,m; int d,b; while(cin>>n>>m && (n||m)){ for(d=sqrt(2.0 *m); d>0;d--){ b=m-(d+d*d)/2; if(b%d==0) cout<<"["<<(b/d)+1<<","<<(b/d)+d<<"]"<<endl; } cout<<endl; } return 0; }
#include <stdio.h> #include <math.h> int main() { int n, m, len, a; while(~scanf("%d%d", &n,&m)) { if(!n && !m) break; len = sqrt(double(m * 2)) + 1; //根据等差数列来算,数列长度 // m = (a + a + len - 1) * len / 2 // m = a * len + len(len - 1)/ 2 // m - len * (len - 1) / 2 = a * len //a = m / len - (len - 1) / 2 while(--len) { a = m / len - (len - 1) / 2; if((a + a + len - 1) * len / 2 == m) { printf("[%d,%d]\n", a, a+len - 1); } } printf("\n"); } return 0; }