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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#380863 | #8049. Equal Sums | ucup-team2230# | TL | 0ms | 7340kb | C++23 | 3.0kb | 2024-04-07 13:52:52 | 2024-04-07 13:52:53 |
Judging History
answer
#ifndef LOCAL
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#endif
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using uint=unsigned;
#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(),x.end()
#define si(x) int(x.size())
#define a first
#define b second
template<class t>using vc=vector<t>;
template<class t,class u>bool chmax(t&a, u b){ if(a<b){a=b;return 1;}return 0;}
template<class t,class u>bool chmin(t&a,u b){if(a>b){a=b;return 1;}return 0;}
using P=pair<int,int>;
const uint mod=998244353;
struct mint{
uint v;
mint(ll vv=0){s(vv%mod+mod);}
mint& s(uint vv){
v=vv<mod?vv:vv-mod;
return *this;
}
mint operator-()const{return mint()-*this;}
mint&operator+=(mint r){return s(v+r.v);}
mint&operator-=(mint r){return s(v+mod-r.v);}
mint&operator*=(mint r){v=(unsigned long long)v*r.v%mod; return *this;}
mint&operator/=(mint r){return *this*=r.inv();}
mint operator+(mint r)const{return mint(*this)+=r;}
mint operator-(mint r)const{return mint(*this)-=r;}
mint operator*(mint r)const{return mint(*this)*=r;}
mint operator/(mint r)const{return mint(*this)/=r;}
mint pow(ll n)const{
if(n<0)return inv(),pow(-n);
mint res(1),x(*this);
while(n){
if(n&1)res*=x;
x*=x;
n>>=1;
}
return res;
}
mint inv()const{return pow(mod-2);}
};
void solve(){
int n,m;
cin>>n>>m;
vc<P> A(n),C(m);
for(int i=0;i<n;i++) cin>>A[i].a>>A[i].b;
for(int i=0;i<m;i++){
cin>>C[i].b>>C[i].a;
C[i].a*=-1;
C[i].b*=-1;
}
int M=505;
int B=2*M*M;
vc<mint> dp(2*B);
dp[B]=1;
vc<vc<mint>> ans(n,vc<mint>(m));
auto add=[&](int l,int r,P p){
//cerr<<l<<" "<<r<<" "<<p.a<<" "<<p.b<<endl;
mint sum=0;
if(p.a<0){
for(int i=l-p.b;i<=l-p.a;i++) sum+=dp[i];
for(int i=l;i<r;i++){
dp[i]=sum;
sum-=dp[i-p.b];
sum+=dp[i-p.a+1];
}
} else{
for(int i=r-p.b;i<=r-p.a;i++) sum+=dp[i];
for(int i=r-1;i>=l;i--){
sum+=dp[i-p.b];
sum-=dp[i-p.a+1];
dp[i]=sum;
}
}
//for(int i=B-5;i<=B+5;i++) cout<<dp[i].v<<" ";cout<<endl;
};
auto rec=[&](int a,int b,int c,int d,auto self)->void{
//cerr<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
if(a==b||c==d) return;
int L=(b-a)+(d-c);
int l=B-L*M,r=B+L*M;
vc<mint> cur(2*L*M);
for(int i=l;i<r;i++) cur[i-l]=dp[i];
if(a+1==b&&c+1==d){
//for(int i=B-5;i<=B+5;i++) cout<<dp[i].v<<" ";cout<<endl;
add(l,r,A[a]);
add(l,r,C[c]);
ans[a][c]=dp[B];
} else{
int ma=(a+b)/2,mc=(c+d)/2;
self(a,ma,c,mc,self);
for(int i=a;i<ma;i++) add(l,r,A[i]);
self(ma,b,c,mc,self);
for(int i=c;i<mc;i++) add(l,r,C[i]);
self(ma,b,mc,d,self);
for(int i=l;i<r;i++) dp[i]=cur[i-l];
for(int i=c;i<mc;i++) add(l,r,C[i]);
self(a,ma,mc,d,self);
}
for(int i=l;i<r;i++) dp[i]=cur[i-l];
};
rec(0,n,0,m,rec);
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cout<<ans[i][j].v<<" ";
}cout<<endl;
}
}
signed main(){
cin.tie(0);
ios::sync_with_stdio(0);
cout<<fixed<<setprecision(20);
solve();
}
详细
Test #1:
score: 100
Accepted
time: 0ms
memory: 7340kb
input:
2 3 1 2 2 3 1 4 2 2 1 3
output:
2 0 0 3 4 4
result:
ok 6 numbers
Test #2:
score: -100
Time Limit Exceeded
input:
500 500 19 458 1 480 7 485 50 461 12 476 15 461 48 466 40 453 46 467 9 458 27 478 26 472 46 459 29 490 6 500 17 487 48 484 28 472 28 459 25 480 4 491 29 481 36 460 2 491 44 499 22 473 20 458 4 483 27 471 2 496 11 461 43 450 2 478 37 466 15 459 42 482 7 451 19 455 2 453 47 475 48 450 1 474 46 471 9 4...