标签:for ++ alt png typedef style info sum 技巧
题目:
思路:
预处理出a[i]在哪个范围区间内是最小的,然后直接遍历a数组求答案就可以了。
这个预处理的技巧巧妙的用了之前的处理结果。(大佬tql)
代码:
#include <bits/stdc++.h> #define inf 0x3f3f3f3f #define MAX 1e3 #define FRE() freopen("in.txt","r",stdin) #define FRO() freopen("out.txt","w",stdout) using namespace std; typedef long long ll; typedef pair<int, int> P; const int maxn = 1000100; int r[maxn],l[maxn]; ll sum[maxn],a[maxn]; int n; int main(){ FRE(); int kase = 0; while(scanf("%d",&n)!=EOF){ sum[0] = 0; a[0] = a[n+1] = -1; for(int i=1; i<=n; i++){ scanf("%lld",&a[i]); sum[i] = sum[i-1]+a[i]; l[i] = r[i] = i; } for(int i=1; i<=n; i++){//根据已经得到的范围快速求出当前的最小值范围 while(a[i] <= a[l[i]-1]){ l[i] = l[l[i]-1]; } } for(int i=n; i>=1; i--){ while(a[i] <= a[r[i]+1]){ r[i] = r[r[i]+1]; } } int L=1,R=1;//当不知道具体的边界的时候,就将边界设为开头 ll ans = a[1]*a[1]; for(int i=1; i<=n; i++){ ll tsum = sum[r[i]] - sum[l[i]-1]; if(ans < tsum * a[i]){ L = l[i]; R = r[i]; ans = tsum*a[i]; } } if(kase++){ printf("\n"); } printf("%lld\n",ans); printf("%d %d\n",L,R); } return 0; }
标签:for ++ alt png typedef style info sum 技巧
原文地址:https://www.cnblogs.com/sykline/p/10352836.html