Feel Good
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 8041 | Accepted: 2177 | |
| 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 10 6 - 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
63 1 6 4 5 2
Sample Output
60
运用单调队列，求出每个数值最左边和最右边能够达到的范围，满足该数值为最小即可。
实现不必像我下面多弄出一个循环，只需一次循环即可
每次元素入栈时，就可更新该元素的范围左端点
每次元素出栈时，就可更新该元素的范围右端点
#include#include #include using namespace std;int a[100005],l[100005],r[100005];long long sum[100005];struct Type{ int res,data;};int main(){ int n; scanf("%d",&n); for(int i=1; i<=n; i++){ scanf("%d",&a[i]); sum[i]=sum[i-1]+a[i]; } deque p; p.clear(); for(int i=1; i<=n; i++){ while(!p.empty()&&p.back().data>=a[i]) p.pop_back(); if(p.empty()) l[i]=1; else l[i]=p.back().res+1; Type t; t.res=i; t.data=a[i]; p.push_back(t); } p.clear(); for(int i=n; i>=1; i--){ while(!p.empty()&&p.back().data>=a[i]) p.pop_back(); if(p.empty()) r[i]=n; else r[i]=p.back().res-1; Type t; t.res=i; t.data=a[i]; p.push_back(t); }// for(int i=1; i<=n; i++)// printf("%d %d\n",l[i],r[i]); long long ans=0; int ansi; for(int i=1; i<=n; i++) if(ans<=(sum[r[i]]-sum[l[i]-1])*a[i]){ ans=(sum[r[i]]-sum[l[i]-1])*a[i]; ansi=i; } printf("%I64d\n",ans); printf("%d %d\n",l[ansi],r[ansi]); return 0;}