【DP】 HDOJ 2993 MAX Average Problem

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n^2的简单DP，用单调队列优化掉n即可。。。
#include <iostream>  #include <queue>  #include <stack>  #include <map>  #include <set>  #include <bitset>  #include <cstdio>  #include <algorithm>  #include <cstring>  #include <climits>  #include <cstdlib>#include <cmath>#include <time.h>#define maxn 100005#define maxm 400005#define eps 1e-10#define mod 1000000007#define INF 999999999#define lowbit(x) (x&(-x))#define mp mark_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid  #define rson o<<1 | 1, mid+1, R  //typedef vector<int>::iterator IT;typedef long long LL;//typedef int LL;using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}// headint q[maxn], sum[maxn], n, num[maxn], k;void read(void){sum[0] = 0;for(int i = 1; i <= n; i++) scanf(num[i]);for(int i = 1; i <= n; i++) sum[i] = sum[i-1] + num[i];}bool check(int a, int b, int c){LL t1 = (LL)(sum[b] - sum[a]) * (c - b);LL t2 = (LL)(sum[c] - sum[b]) * (b - a);if(t1 > t2) return true;else return false;}bool solve(int a, int b, int c){LL t1 = (LL)(sum[c] - sum[b]) * (c - a);LL t2 = (LL)(sum[c] - sum[a]) * (c - b);if(t1 > t2) return true;else return false;}void work(void){ int l = 0, r = 0; double t = 0, mx = 0; for(int i = 0; i + k <= n; i++) { while(r - l >= 2 && i <= n - k && check(q[r-2], q[r-1], i)) r--; q[r++] = i; while(r - l >= 2 && solve(q[l], q[l+1], i + k)) l++; t = 1.0 * (sum[i + k] - sum[q[l]]) / (i + k - q[l]); mx = max(mx, t); } printf("%.2f\n", mx);}int main(void){while(scanf("%d%d", &n, &k)!=EOF) {read();work();}return 0;}
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