nowcoder20 A 柯朵莉

Posted 2020-10-14 2855669158

题意:一个序列长度为n,m次询问每次询问一个x，要求把这个序列分成每段不超过x，问分的段数的期望

题解：好久没做期望了，都忘了。。。期望倒着算，dp[x]代表X后缀的期望，dp[n] = 1，只要双指针处理出来每个数能到的最有的下标，后缀保存一下，倒着算就可以了

#include <bits/stdc++.h>
#define maxn 101000
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
int a[maxn], x, n, m, l[maxn], sum;
double dp[maxn], cnt[maxn];
int main(){
    int ma = 0;
    scanf("%d%d", &n, &m);
    for(int i=1;i<=n;i++){
        scanf("%d", &a[i]);
        ma = max(a[i], ma);
    }
    while(m--){
        scanf("%d", &x);
        if(x<ma){
            printf("YNOI is good OI!\n");
            continue;
        }
        sum = 0;
        int i=1,j=1;
        for(i=1,j=1;i<=n&&j<=n;i++){
            while(sum+a[j]<=x&&j<=n) sum += a[j], j++;
            l[i] = j-1;
            sum -= a[i];
        }
        for(int k=i;k<=n;k++) l[k] = n;
        cnt[n] = 1;
        for(int i=n-1;i>=1;i--){
            dp[i] = (cnt[i+1]-cnt[l[i]+2])*1.0/(l[i]-i+1)+1;
            cnt[i] = cnt[i+1]+dp[i];
        }
        printf("%.2f\n", dp[1]);
    }
    return 0;
}