蒟蒻DP专题训练2--HDU1231
2017-05-08 17:24
393 查看
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
);
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;
}
#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
);
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;
}
相关文章推荐
- dp专题训练
- 线段树专题#3_蒟蒻训练历程记录_HDU1754_I hate It
- 线段树专题#4_蒟蒻训练历程记录_HDU1698_ 延迟标记、区间更新
- 线段树专题#6_蒟蒻训练历程记录_HDU 2705 Billboard_单点更新
- 【蒟蒻的点分治专题训练】----5道题题解
- 专题(弱点)Dp训练总结【状压Dp*1+区间Dp*5+数位dp*3+树型Dp*2】【10/11】
- 2016 SCUT 专题训练 简单dp
- dp暑假专题 训练记录
- 概率dp && 高斯消元 专题训练
- 11/12训练日记 数位dp专题结束
- ACdream DP专题训练
- 线段树专题#1_蒟蒻训练历程记录_HDU1166
- 线段树专题#5_蒟蒻训练历程记录_HDU 1394 Minimum Inversion Number_单点更新+思维转换
- 暑假训练2-DP专题
- 线段树专题#2_蒟蒻训练历程记录_HDU5775
- HFUT&AHU组团训练(一)----DP专题
- HDOJ树形DP专题之Anniversary party
- 周中训练笔记19——树形DP总结
- DP专题 DP46
- 暑假训练专题一 博弈