wenbao与二分加贪心

 

 

https://vjudge.net/contest/144453#problem/A

　　长城守卫

 

 

 1 #include <iostream>
 2 #include <stdio.h>
 3 using namespace std;
 4 const int maxn = 1e5+10;
 5 int n, a[maxn], le[maxn], ri[maxn];
 6 int Max(int x, int y) {return x > y ? x : y; }
 7 int Min(int x, int y) {return x < y ? x : y; }
 8 bool ok(int x){
 9     le[1] = a[1], ri[1] = 0;
10     int y = x - a[1];
11     for(int i = 2; i <= n; i++){
12         if(i&1){
13             ri[i] = Min(a[i], y - ri[i-1]);
14             le[i] = a[i] - ri[i];
15         }else{
16             le[i] = Min(a[i], a[1] - le[i-1]);
17             ri[i] = a[i] - le[i];
18         }
19     }
20     if(le[n] == 0) return true;
21     else return false;
22 }
23 int main(){
24     while(scanf("%d", &n)&&n){
25         for(int i = 1; i <= n; i++) scanf("%d", a+i);
26         int L = -1, R = -1;
27         a[n+1] = a[1];
28         for(int i = 1; i <= n; i++){
29             L = Max(L, a[i]+a[i+1]), R = Max(R, a[i]*3);
30         }
31         if(n == 1) {
32             printf("%d\n", a[1]);
33             continue;
34         }
35         if(n&1){
36             while(L < R){
37                 int mid = (L+R) >> 1;
38                 if(ok(mid)){
39                     R = mid;
40                 }else{
41                     L = mid + 1;
42                 }
43             }
44         }
45         printf("%d\n", L);
46     }
47     return 0;
48 }

 

 

 

 

只有不断学习才能进步！

posted @ 2018-04-14 13:52  wenbao  阅读(139)  评论(0)  编辑  收藏  举报