链接： http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=2116
Description
Wind and his GF(game friend) are playing a small game. They use the computer to randomly generated a number sequence which only include number 2,0 and -2. To be the winner,wind must
have to find a continuous subsequence whose continuous product is maximum.
For example, we have a sequence blow:
2 2 0 -2 0 2 2 -2 -2 0
Among all of it‘s continuous subsequences, 2 2 -2 -2 own the maximum continuous product.
(2*2*(-2)*(-2) = 16 ,and 16 is the maximum continuous product)
You,wind‘s friend,can give him a hand.
Input
The first line is an integer T which is the Case number(T <= 200).
For each test case, there is an integer N indicating the length of the number sequence.（1<= N <= 10000)
The next line,there are N integers which only include 2,0 and -2.
Output
For each case,you have to output the case number first(Reference the sample).
If the answer is smaller than 0, you just need to output 0 as the answer.
If the answer‘s format is 2^x,you need to output the x as the answer.
Output the answer in one line.
Sample Input
2
2
-2 0
10
2 2 0 -2 0 2 2 -2 -2 0
Sample Output
Case #1: 0
Case #2: 4
代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstring>
#define MAXN 15000
#define RST(N)memset(N, 0, sizeof(N))
using namespace std;
int n, ans, cas, T = 1;
int a[MAXN], mn[MAXN], mx[MAXN];
void Init()
{
scanf("%d", &n);
for(int i=1; i<=n; i++) scanf("%d", &a[i]);
RST(mx), RST(mn), ans = 0;
}
int solve()
{
for(int i=1; i<=n; i++) {
if(a[i] > 0) {
if(mn[i-1] > 0) mn[i] = mn[i-1]+1;
mx[i] = mx[i-1]+1;
}else if(a[i] < 0) {
if(mn[i-1] > 0) mx[i] = mn[i-1]+1;
if(mx[i-1] > 0) mn[i] = mx[i-1]+1;
else mn[i] = 1;
}else {
mx[i] = 0;
mn[i] = 0;
}
ans = max(ans, mx[i]);
}
return ans;
}
int main()
{
scanf("%d", &cas);
while(cas--) {
Init();
printf("Case #%d: %d\n", T++, solve());
}
return 0;
}
HLG 2116 Maximum continuous product （最大连续积 DP）