UVa11059 Maximum Product

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题意求最大连续子序列的积

思路本题可以暴力枚举起始和终止位置，O(n^2)，不过为了练习，写了一个O(n)的算法，总体来算法不过简练

大致思路，就是正反两个方向遍历数组，预处理对所有负数的位置，保存在下一个0前还有几个负数，在上一个0后还有几个负数，然后双向遍历，遇到正数，更新sum，遇到0，清空sum，更新max，遇到负数，要判断若当前sum是负数或者该负数后，在下一个0前还有负数，则更新sum，否则清空sum。

#include <iostream>#include <cstdio>#include <cstring>#include <map>#include <vector>#include <algorithm>using namespace std;#define MAX 20typedef long long ll;typedef pair<int,int> Pll;#define fi first#define se second#define MP make_pairint n;ll max1 = 0;ll a[MAX];Pll b[MAX];int main(){int i,j;int T = 0;while(scanf("%d",&n) == 1){for(i=0;i<n;i++)scanf("%lld",&a[i]);T++;max1 = 0;int count = 0;for(i=n-1;i>=0;i--){if(a[i] < 0)b[i].fi = count++;else if(a[i] == 0)count = 0;}count = 0;for(i=0;i<n;i++){if(a[i] < 0)b[i].se = count++;else if(a[i] == 0)count = 0;}ll sum = 1;bool flag = 0;for(i=0;i<n;i++){if(a[i] > 0){sum *= a[i];flag = 1;}else if(a[i] == 0){max1 = max(sum,max1);sum = 1;}else{if(sum < 0 || b[i].fi > 0){sum *= a[i];flag = 1;}else{max1 = max(sum,max1);sum = 1;}}}max1 = max(sum,max1);sum = 1;for(i=n-1;i>=0;i--){if(a[i] > 0){sum *= a[i];flag = 1;}else if(a[i] == 0){max1 = max(sum,max1);sum = 1;}else{if(sum < 0 || b[i].se > 0){sum *= a[i];flag = 1;}else{max1 = max(sum,max1);sum = 1;}}}max1 = max(sum,max1);if(flag == 0)printf("Case #%d: The maximum product is 0.\n\n",T);elseprintf("Case #%d: The maximum product is %lld.\n\n",T,max1);}return 0;}

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