题意:
最低等级(level 1),已知在(level i)操作一次需花费(a_i),有概率(p_i)升级到(level i+1),有(1 - p_i)掉级到(x_i(x_i <= i)),询问(q)次,问你每次从(l)升级到(r)的花费的期望。
思路:
我们设(dp[i])为从(1)升级到(i)的期望花费,那么显然有从(l)升级到(r)的期望花费为(dp[r] - dp[l])。
然后我们可以知道,升级到(i)有两种情况:
已经花费了(dp[i-1])必加。从(i-1)升级,那么花费(a_{i-1});掉级到(x_{i-1})再升到(i),那么花费(a_{i-1} + dp[i]-dp[x_{i-1}]),那么可以列出方程:
[dp[i] = dp[i -1] + p_{i-1}*a_{i-1} + (1-p_{i-1})*(dp[i]-dp[x_{i-1}]+a_{i-1})
]
化简后可得正解。
代码:
#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 5e5 + 5;
const ll MOD = 1e9+7;
const ull seed = 13131;
const int INF = 998244353;
ll dp[maxn];
ll r[maxn], s[maxn], x[maxn], a[maxn];
ll ppow(ll a, ll b){
ll ret = 1;
while(b){
if(b & 1) ret = ret * a % MOD;
b >>= 1;
a = a * a % MOD;
}
return ret;
}
int main(){
int T;
scanf("%d", &T);
while(T--){
int n, q;
scanf("%d%d", &n, &q);
for(int i = 1; i <= n; i++){
scanf("%lld%lld%lld%lld", &r[i], &s[i], &x[i], &a[i]);
}
dp[1] = 0;
for(int i = 2; i <= n + 1; i++){
ll inv = ppow(r[i - 1], MOD - 2);
ll tmp1 = ((s[i - 1] * inv % MOD) * dp[i - 1] + a[i - 1]) % MOD;
ll tmp2 = ((s[i - 1] - r[i - 1]) * inv % MOD) * dp[x[i - 1]] % MOD;
ll tmp3 = ((s[i - 1] - r[i - 1]) * inv % MOD) * a[i - 1] % MOD;
dp[i] = ((tmp1 - tmp2) % MOD + tmp3) % MOD;
dp[i] = (dp[i] % MOD + MOD) % MOD;
}
// for(int i = 1; i <= n + 1; i++) printf("%lld
", dp[i]);
while(q--){
int l, r;
scanf("%d%d", &l, &r);
ll ans = dp[r] - dp[l];
ans = (ans % MOD + MOD) % MOD;
printf("%lld
", ans);
}
}
return 0;
}