Kejin Player 期望DP
题意：
初始等级为1，每一级有四个参数 r , s , x , a 。
每一级有一个概率p=r/s花费a的代价升级到下一级，失败可能会倒退到x级
设从 l 到 r 的期望为 g(l, r), 这种期望满足减法 g(l, r) = g(1, r) − g(1, l)．
因为升级只能一级一 级升, 所以要从 1 升级到 r, 必然要经过 l．
求一个前缀和
sum[i+1]=sum[i]+ai * si / ri + (sum[i]-sum[x])*(si-ri)/ri;
ai * si / ri + (sum[i]-sum[x])*(si-ri)/ri 这个类似于二项分布
一个是升到下一级的花费，一个是失败后回到 i 这一级的花费
1 #include <set> 2 #include <map> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <cstdio> 7 #include <string> 8 #include <vector> 9 #include <time.h> 10 #include <cstring> 11 #include <iostream> 12 #include <algorithm> 13 #include <unordered_map> 14 15 16 #define pi acos(-1.0) 17 #define eps 1e-9 18 #define fi first 19 #define se second 20 #define rtl rt<<1 21 #define rtr rt<<1|1 22 #define bug printf("******\n") 23 #define mem(a, b) memset(a,b,sizeof(a)) 24 #define name2str(x) #x 25 #define fuck(x) cout<<#x" = "<<x<<endl 26 #define sf(n) scanf("%d", &n) 27 #define sff(a, b) scanf("%d %d", &a, &b) 28 #define sfff(a, b, c) scanf("%d %d %d", &a, &b, &c) 29 #define sffff(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d) 30 #define pf printf 31 #define FIN freopen("../date.txt","r",stdin) 32 #define gcd(a, b) __gcd(a,b) 33 #define lowbit(x) x&-x 34 #define IO iOS::sync_with_stdio(false) 35 36 37 using namespace std; 38 typedef long long LL; 39 typedef unsigned long long ULL; 40 const int maxn = 1e6 + 7; 41 const int maxm = 8e6 + 10; 42 const int INF = 0x3f3f3f3f; 43 const int mod = 1e9 + 7; 44 45 LL expmod(LL a, LL b) { 46 LL res = 1; 47 while (b) { 48 if (b & 1) res = res * a % mod; 49 a = a * a % mod; 50 b = b >> 1; 51 } 52 return res; 53 } 54 55 56 int t, n, q; 57 struct node { 58 LL r, s, x, a; 59 } qu[maxn]; 60 LL sum[maxn]; 61 62 int main() { 63 //FIN; 64 sf(t); 65 while (t--) { 66 sff(n, q); 67 for (int i = 1; i <= n; i++) scanf("%lld%lld%lld%lld", &qu[i].r, &qu[i].s, &qu[i].x, &qu[i].a); 68 sum[1] = 0; 69 for (int i = 1; i <= n; i++) { 70 sum[i + 1] = sum[i] + qu[i].s * expmod(qu[i].r, mod - 2) % mod * qu[i].a % mod + 71 (qu[i].s - qu[i].r + mod) % mod * expmod(qu[i].r, mod - 2) % mod * 72 (sum[i] - sum[qu[i].x] + mod) % mod; 73 // printf("sum[%d] = %d\n", i + 1, sum[i + 1]); 74 sum[i+1]%=mod; 75 } 76 while (q--) { 77 int L, R; 78 sff(L, R); 79 printf("%lld\n", (sum[R] - sum[L]+mod)%mod); 80 } 81 } 82 return 0; 83 }