【POJ 2796】 Feel Good
2016-04-14 23:50
411 查看
Feel Good
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 12134 Accepted: 3376
Case Time Limit: 1000MS Special Judge
Description
Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people’s memories about some period of life.
A new idea Bill has recently developed assigns a non-negative integer value to each day of human life.
Bill calls this value the emotional value of the day. The greater the emotional value is, the better the daywas. Bill suggests that the value of some period of human life is proportional to the sum of the emotional values of the days in the given period, multiplied by the smallest emotional value of the day in it. This schema reflects that good on average period can be greatly spoiled by one very bad day.
Now Bill is planning to investigate his own life and find the period of his life that had the greatest value. Help him to do so.
Input
The first line of the input contains n - the number of days of Bill’s life he is planning to investigate(1 <= n <= 100 000). The rest of the file contains n integer numbers a1, a2, … an ranging from 0 to 106 - the emotional values of the days. Numbers are separated by spaces and/or line breaks.
Output
Print the greatest value of some period of Bill’s life in the first line. And on the second line print two numbers l and r such that the period from l-th to r-th day of Bill’s life(inclusive) has the greatest possible value. If there are multiple periods with the greatest possible value,then print any one of them.
Sample Input
6
3 1 6 4 5 2
Sample Output
60
3 5
Source
Northeastern Europe 2005
**【题解】【第一次写单调栈……其实和单调队列还是蛮像的】
【这个题,是要求一个递增区间,输出的是这个区间的和乘这个区间里的最小值,要求这个最小值最大,并输出这个区间的左右端点】
【首先,在读入的时候,要初始化前缀和还有包含当前点的区间的左右端点,最开始的时候,左右端点是当前点。然后运用单调栈来处理 】
【维护一个递增的栈,每次向里面插元素的时候,先将所有大于当前点的值全部出栈,在出栈前要伸展区间的左右端点(因为维护的是递增的序列,所以出栈的数其实都可以包含在当前区间)。 再将所有元素都处理完后,将所有元素都出栈,并同时再度伸展每个区间的左右端点,以防有没有更新到位的】
【在输出时,用前缀和来寻找递增区间和乘区间最小值的最大值】**
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 12134 Accepted: 3376
Case Time Limit: 1000MS Special Judge
Description
Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people’s memories about some period of life.
A new idea Bill has recently developed assigns a non-negative integer value to each day of human life.
Bill calls this value the emotional value of the day. The greater the emotional value is, the better the daywas. Bill suggests that the value of some period of human life is proportional to the sum of the emotional values of the days in the given period, multiplied by the smallest emotional value of the day in it. This schema reflects that good on average period can be greatly spoiled by one very bad day.
Now Bill is planning to investigate his own life and find the period of his life that had the greatest value. Help him to do so.
Input
The first line of the input contains n - the number of days of Bill’s life he is planning to investigate(1 <= n <= 100 000). The rest of the file contains n integer numbers a1, a2, … an ranging from 0 to 106 - the emotional values of the days. Numbers are separated by spaces and/or line breaks.
Output
Print the greatest value of some period of Bill’s life in the first line. And on the second line print two numbers l and r such that the period from l-th to r-th day of Bill’s life(inclusive) has the greatest possible value. If there are multiple periods with the greatest possible value,then print any one of them.
Sample Input
6
3 1 6 4 5 2
Sample Output
60
3 5
Source
Northeastern Europe 2005
**【题解】【第一次写单调栈……其实和单调队列还是蛮像的】
【这个题,是要求一个递增区间,输出的是这个区间的和乘这个区间里的最小值,要求这个最小值最大,并输出这个区间的左右端点】
【首先,在读入的时候,要初始化前缀和还有包含当前点的区间的左右端点,最开始的时候,左右端点是当前点。然后运用单调栈来处理 】
【维护一个递增的栈,每次向里面插元素的时候,先将所有大于当前点的值全部出栈,在出栈前要伸展区间的左右端点(因为维护的是递增的序列,所以出栈的数其实都可以包含在当前区间)。 再将所有元素都处理完后,将所有元素都出栈,并同时再度伸展每个区间的左右端点,以防有没有更新到位的】
【在输出时,用前缀和来寻找递增区间和乘区间最小值的最大值】**
#include<cstdio> #include<cstring> #include<algorithm> #define ll long long using namespace std; struct zhan{ ll l,r; }a[100010]; ll n,tot,stack[100010],d[100010],sum[100010],maxn,ansl,ansr; inline void init(ll x) { if (x==1) {stack[++tot]=x; return;} while(d[stack[tot]]>=d[x]&&tot) { a[x].l=a[stack[tot]].l; a[stack[tot-1]].r=a[stack[tot]].r; tot--; } a[stack[tot]].r=a[x].r; stack[++tot]=x; return; } inline void pop() { a[stack[tot-1]].r=a[stack[tot]].r; tot--; } int main() { ll i,j; scanf("%lld",&n); for (i=1;i<=n;i++) { scanf("%lld",&d[i]); sum[i]=sum[i-1]+d[i]; a[i].l=a[i].r=i; } for(i=1;i<=n;++i) init(i); while(tot) pop(); for(i=1;i<=n;++i) { ll s1=sum[a[i].r]-sum[a[i].l-1]; if (maxn<=s1*d[i]) maxn=s1*d[i],ansl=a[i].l,ansr=a[i].r; } printf("%lld\n",maxn); printf("%lld %lld\n",ansl,ansr); return 0; }
相关文章推荐
- JavaScript函数劫持
- 在线压缩javascript --- Google Closure Compiler
- React one-way flow
- 用VueJS+webpack+semantic-UI+Laravel开发单页应用
- 二次贝塞尔曲线的实现(基于Canvas与JavaScript)
- css 伪元素的初次使用
- jstat监控JVM内存使用情况、GC回收情况
- 【JavaScript】JS语言原生支持Hash
- 看着看着就哭了的前端地址大全
- jQuery图片预加载
- javascript 理解闭包
- 几种延迟加载JS代码的方法加快网页的访问速度
- AngularJs学习——模拟用户登录的简单操作
- AngularJS学习笔记(一) 关于MVVM和双向绑定
- JS 速度动画
- jquery 添加或删除HTML属性
- 原型(3) JavaScript原型链详细介绍
- html 之input标签height设置问题
- jquery.cookie.js使用
- 10段代码打通js学习的任督二脉