CF1009E Intercity Travelling

Posted 2021-08-29 Jozky86

CF1009E Intercity Travelling

题意：

有一段路程，路程可以看作是从0到n的一条直线
如果从起点出发或者从休息点出发，连续驾驶k千米，则需要消耗的体能为a1+…+ak
每个整点都有可能拥有一个休息点，每个休息点存在或者不存在的概率想等的。
求整个旅程消耗的体能的期望为p，输出$p\ast 2^{n-1}$

题解：

最后别忘乘$2^n-1$

代码：

#include <bits/stdc++.h>
#define LL long long
#define MAXN 1000005
#define P int(998244353)
using namespace std;

int N;
LL a[MAXN], s[MAXN];
LL pw[MAXN];
LL dp[MAXN], s1[MAXN], s2[MAXN];

/*
dp[i] = (s[i] + sum{(dp[j]+s[i-j])*2^(j-1)})/2^(i-1), j = 1...i-1
      = (s[i] + sum{dp[j]*2^(j-1)} + 2^(i-1)*sum{s[i-j]/2^(i-j)})/2^(i-1)
      = s[i]/2^(i-1) + sum{dp[j]*2^j}/2^i + sum{s[j]/2^j}, j=1...i-1
*/

LL qpow(LL x, int n){
    LL res = 1;
    while(n){
        if(n & 1) res = res * x % P;
        x = x * x % P;
        n /= 2;
    }
    return res;
}

int main(){

    scanf("%d", &N);

    pw[0] = 1;
    for(int i=1;i<=N;i++){
        scanf("%d", &a[i]);
        s[i] = (s[i-1] + a[i]) % P;
        pw[i] = pw[i-1] * 2 % P;
    }

    for(int i=1;i<=N;i++){
        dp[i] = (s[i]*qpow(pw[i-1], P-2)%P + s1[i-1]*qpow(pw[i], P-2)%P + s2[i-1]%P)%P;

        s1[i] = (s1[i-1] + dp[i]*pw[i]%P)%P;
        s2[i] = (s2[i-1] + s[i]*qpow(pw[i],P-2)%P)%P;
    }

    printf("%lld\\n", dp[N]*pw[N-1]%P);

    return 0;
}