The sum problem
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15180 Accepted Submission(s): 4552
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]
#include#include #define ll __int64 int main() { ll n,m; ll i,k; while(scanf("%I64d%I64d",&n,&m),n||m) { for(k=sqrt((double)2*m);k>=1;k--)//(a+a+k-1)*k/2<=m a>=1所以(k*k+k)<=2*m 但是因为输出格式问题. 必须先输出间隔长的. 所以就多算了一点.把(k*k+k)<=2*m 当做了 k*k<=2*m { double tem=(2.0*m/k+1-k)/2.0; if(tem==(ll)tem&&(ll)tem+k-1<=n&&(ll)tem>=1) { printf("[%I64d,%I64d]\n",(ll)tem,(ll)tem+k-1); } } puts(""); } return 0; }