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]
数学思维题目，，看了一下大佬的博文。。感觉 还复杂啊！！
#include<iostream> #include<algorithm> #include<string> #include<cstdio> #include<math.h> using namespace std; /* 思路: 区间项数为i 起始位置 a,终点为b则b=a+i-1; 这一段区间的和为sum=(a+b)*i/2=(a+a+i-1)*i/2=m; a最小为1，则（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);判断条件就是 (a+a+i-1)*i==2*m; */ int main() { int n,m,a,b; while(cin>>n>>m) { if(n==0 && m==0) break; for(int i=sqrt(2*m);i>=1;i--)//最小为1项，就是[m,m] { a=m/i-(i-1)/2;//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-1)*i==2*m) { printf("[%d,%d]\n",a,a+i-1); } } printf("\n"); } return 0; }
原文：https://www.cnblogs.com/Accepting/p/11208440.html