1 solutions
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0
#include <stdio.h> int main() { int n,m,k,j,f,a,b,d; int i = 1; int sum = 0; scanf("%d", &n); scanf("%d", &m); k = n / m; int num[1000]; int sum1[1000] = { 0 }; int sum2[1000] = { 0 }; for (i = 0; i < n; i++) { scanf("%d", &num[i]); } for (i = 0; i < m; i++) { for (j = i * k; j < k * (i + 1); j++) sum1[i] += num[j]; } for (i = 0; i < m; i++) { printf("%d ", sum1[i]); } printf("\n"); for (i = 2;( m / i) != 1; i=i*2) { for (j = 0; j < m / i; j++) { for (a = 0; a < i * (j + 1); a++) sum2[j] += sum1[a]; } for (f = (m / i) - 1; f > 0; f--) sum2[f] = sum2[f] - sum2[f - 1]; for (d = 0; d < m / i; d++) printf("%d ", sum2[d]); printf("\n"); for (b = 0; b < m; b++) { sum2[b] = 0; } } for (i = 0; i < m; i++) sum += sum1[i]; printf("%d",sum); return 0; }
- 1
Information
- ID
- 6660
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 7
- Tags
- (None)
- # Submissions
- 63
- Accepted
- 17
- Uploaded By