hdu 2058 The sum problem
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The sum problem
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20505 Accepted Submission(s): 6023
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]
解析:直接枚举会超时,因此枚举时要缩小范围。根据等差求和公式Sn = n*a1+n*(n-1)/2*d,
此处Sn = m,d = 1,则m = n*a1+n*(n-1)/2。整理得2*m = n*n+(2*a1-1)*n。可得:
n<sqrt(2.0*m)。
1 #include <cstdio> 2 #include <cmath> 3 4 int n,m; 5 6 int main() 7 { 8 while(scanf("%d%d",&n,&m), n){ 9 for(int num = (int)sqrt(m*2.0); num>0; --num){ 10 int numa1 = m-num*(num-1)/2; 11 if(numa1%num == 0) 12 printf("[%d,%d]\n",numa1/num,numa1/num+num-1); 13 } 14 printf("\n"); 15 } 16 return 0; 17 }
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HDU 2058 The sum problem (数学+暴力)