2019 Multi-University Training Contest 2

http://acm.hdu.edu.cn/search.php?action=listproblem

Kejin Player

思路：ri/si为等级为i时升级成功的概率，si/ri就表示第si/ri次升级成功，而前面的（si/ri-1）次全部升级失败，用这样的方式计算期望。

#include<bits/stdc++.h>
using namespace std;

#define ll long long
const int mod=1e9+7;
const int maxn=5e5+10;
ll power(ll a, ll b, ll p)
{
    ll ans=1%p;
    for(; b; b>>=1)
    {
        if(b&1) ans=(long long)ans*a%p;
        a=(long long)a*a%p;
    }
    return ans;
}
ll sum[maxn];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int tt;
    cin>>tt;
    while(tt--)
    {
        ll n,q;
        cin>>n>>q;
        ll r,s,x,a;
        sum[1]=0;
        for(int i=1; i<=n; i++)
        {
            cin>>r>>s>>x>>a;
            ll t=s*power(r,mod-2,mod)%mod;
            sum[i+1]=(sum[i]+(ll)t*a%mod+((ll)(t-1+mod)*((sum[i]-sum[x]+mod)%mod)%mod))%mod;
            
        }
        for(int i=1; i<=q; i++)
        {
            ll l,r;
            cin>>l>>r;
            cout<<(sum[r]-sum[l]+mod)%mod<<endl;
        }
    }
}