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#include <stdio.h> void MaxSubsseqSum1(int A[], int N) { int ThisSum = 0, MaxSum = -1; int start,end; int temp = 0; for (int i = 0; i<N; i++) { ThisSum += A[i]; if (ThisSum>MaxSum){ MaxSum = ThisSum; start = temp; end = i; } else if (ThisSum<0) { ThisSum = 0; temp = i+1; } } if (MaxSum>=0) { printf("%d %d %d\n",MaxSum, A[start], A[end]); } else { printf("0 %d %d\n", A[0], A[N-1]); } } int main(void) { int N; scanf("%d",&N); int A[N]; for (int i = 0; i<N;i++) { scanf("%d",&A[i]); } MaxSubsseqSum1(A, N); }
01-复杂度2 Maximum Subsequence Sum (25分)
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原文地址:http://www.cnblogs.com/sjdeak/p/4781841.html