Submission #1020149

#	Time	Username	Problem	Language	Result	Execution time	Memory
1020149		bachhoangxuan	Trains (BOI24_trains)	C++17	100 / 100	228 ms	280408 KiB

Main

// Judges with GCC >= 12 only needs Ofast
// #pragma GCC optimize("O3,no-stack-protector,fast-math,unroll-loops,tree-vectorize")
// MLE optimization
// #pragma GCC optimize("conserve-stack")
// Old judges
// #pragma GCC target("sse4.2,popcnt,lzcnt,abm,mmx,fma,bmi,bmi2")
// New judges. Test with assert(__builtin_cpu_supports("avx2"));
// #pragma GCC target("avx2,popcnt,lzcnt,abm,bmi,bmi2,fma,tune=native")
// Atcoder
// #pragma GCC target("avx2,popcnt,lzcnt,abm,bmi,bmi2,fma")
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
- insert(x),erase(x)
- find_by_order(k): return iterator to the k-th smallest element
- order_of_key(x): the number of elements that are strictly smaller
*/
#include<bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
uniform_real_distribution<> pp(0.0,1.0);
#define int long long
#define ld long double
#define pii pair<int,int>
#define piii pair<int,pii>
#define mpp make_pair
#define fi first
#define se second
const long long inf=1e18;
const int mod=1e9+7;
const int maxn=200005;
const int B=350;
const int maxs=355;
const int maxm=200005;
const int maxq=1000005;
const int maxl=25;
const int maxa=1000000;
const int root=3;
int power(int a,int n){
    int res=1;
    while(n){
        if(n&1) res=res*a%mod;
        a=a*a%mod;n>>=1;
    }
    return res;
}
const int iroot=power(3,mod-2);
const int base=101;

int n,x[maxn],d[maxn];
int sum[maxs][maxn],dp[maxn];

void solve(){
    cin >> n;
    for(int i=1;i<=n;i++){
        cin >> d[i] >> x[i];
        if(d[i]) x[i]=min(x[i],(n-i)/d[i]);
    }
    for(int i=n;i>=1;i--){
        dp[i]=1;
        if(d[i]){
            if(d[i]<=B) dp[i]=(dp[i]+sum[d[i]][i+d[i]]-sum[d[i]][i+(x[i]+1)*d[i]]+mod)%mod;
            else{
                for(int j=1;j<=x[i];j++) dp[i]=(dp[i]+dp[i+j*d[i]])%mod;
            }
        }
        for(int j=1;j<=B;j++) sum[j][i]=(dp[i]+sum[j][i+j])%mod;
        //cout << dp[i] << '\n';
    }
    cout << dp[1] << '\n';
}

signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    int test=1;//cin >> test;
    while(test--) solve();
}

• Subtask 1: 8 / 8
• Subtask 2: 13 / 13
• Subtask 3: 16 / 16
• Subtask 4: 34 / 34
• Subtask 5: 29 / 29