蒟蒻DP专题训练2--HDU1231

来源：互联网发布：网络信息有限公司名字编辑：程序博客网时间：2024/06/07 12:23

简单的一维dp，但是怕会超int，一定要开成long long，注意读取和打印都是lld，否则会出错。然后，题目里面说如果全是负数则输出第一个元素和最后一个元素，还是要审题。然后在每次遇到转移情况为重开区间的时候，要更新区间左值和右值，然后在寻找最大值的时候记录一下最大值和相应的区间左右值就好了，一次AC，代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<vector>#include<map>#include<algorithm>#include<queue>#include<stack>using namespace std;typedef long long ll;ll a[10010];ll dp[10010];//dp[i]表示以i为结尾的最大连续子序咧的和 int res[10010];int main(){int n;ll fmax = -0x3f3f3f3f;while(~scanf("%d", &n) && n != 0){int l = 1, r = 1, l1 = 1, r1 = 1, flag = 0;memset(dp, 0, sizeof(dp));memset(res, 0, sizeof(res));for(int i = 1; i <= n; i++){scanf("%lld", &a[i]);if(a[i] >= 0)   flag = 1;}if(!flag){printf("0 %lld %lld\n", a[1], a[n]);continue;}   fmax = a[1];dp[1] = a[1];for(int i = 2; i <= n; i++){if(dp[i-1]+a[i] > a[i]){dp[i] = dp[i-1]+a[i];r = i;if(dp[i] > fmax){fmax = dp[i];l1 = l, r1 = r;}} else{dp[i] = a[i];l = i;r = i;if(dp[i] > fmax){fmax = dp[i];l1 = l, r1 = r;//printf("i:%d dp=%lld l:%d r:%d\n", i, dp[i], l, r);}}//printf("i:%d dp=%lld l:%d r:%d l1:%d r1:%d\n", i, dp[i], l, r, l1, r1);} printf("%lld %lld %lld\n", fmax, a[l1], a[r1]);}return 0;}

0 0