The sum problem
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29632 Accepted Submission(s): 8887
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.
Sample Input
20 10
50 30
0 0
Sample Output
[1,4]
[10,10]
[4,8]
[6,9]
[9,11]
[30,30]
分析：设满足条件的区间为[a,a+k-1],那么M=a+(a+1)+......+(a+k-1)=k*a+k*(k-1)/2,
然后枚举k,算出a,记录结果。
#include<cstdio> long long a[1000000],num[1000000]; int main() { long long N,M; while(scanf("%lld%lld",&N,&M)) { if(!N&&!M) break; int cnt=0; long long k=1,temp; while(true) { temp=M-k*(k-1)/2; if(temp<=0) break; if(temp%k==0&&temp/k+k-1<=N) {num[cnt]=k-1;a[cnt++]=temp/k;} k++; } for(int i=cnt-1;i>=0;i--) printf("[%lld,%lld]\n",a[i],a[i]+num[i]); printf("\n"); } return 0; }
