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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 |
---|---|---|---|---|---|---|---|---|---|
#290807 | #6186. Cryptography Problem | light_ink_dots# | TL | 0ms | 0kb | C++20 | 2.5kb | 2023-12-25 15:51:55 | 2023-12-25 15:51:56 |
answer
#include<bits/stdc++.h>
using namespace std;
const int maxn=2005,S1=200,S2=1000,S3=3,S4=5,S5=300,S6=9;
int T,n;
int p[10];
long long mod,err,nowL,nowR,ans;
long long a[maxn],c[maxn];
struct node{
long long a,c,cnt;
inline bool operator ==(node &p)const{
return a==p.a;
}
inline bool operator <(node &p)const{
return a<p.a;
}
};
vector<node>V;
inline long long inc(long long x){
return x>=mod? x-mod:x;
}
inline long long dec(long long x){
return x<0? x+mod:x;
}
inline long long mul(long long x,long long y){
return (x*y-(long long)((long double)x/mod*y)*mod+mod)%mod;
}
int check(long long X){
X=(X%mod+mod)%mod;
int flg=1;
for(int i=1;i<=n;i++){
long long val=dec(mul(a[i],X)-c[i]);
flg&=(val<=err||val>=mod-err);
}
return flg;
}
mt19937_64 rnd(144441);
void addnode(long long a,long long c,long long cnt){
V.push_back(node{a,c,cnt});
if(err>mod/cnt/2||(nowR-nowL+1)>mod/a)
return ;
long long L=mul(nowL,a),R=mul(nowR,a),tmpL=c-cnt*err,tmpR=c+cnt*err;
while(L>R)
R+=mod;
while(tmpL>R)
tmpL-=mod,tmpR-=mod;
while(tmpR<L)
tmpL+=mod,tmpR+=mod;
if(tmpL+mod>R&&tmpR-mod<L)
nowL+=max(tmpL-L,0ll)/a,nowR-=max(R-tmpR,0ll)/a;
}
int main(){
srand(time(0));
scanf("%d",&T);
while(T--){
scanf("%d%lld",&n,&mod),err=(mod/100+1)/2,nowL=0,nowR=mod-1;
for(int i=1;i<=n;i++)
scanf("%lld%lld",&a[i],&c[i]);
while(nowR-nowL+1>S1){
V.clear();
long long delta=rnd()%mod;
delta=0;
nowL+=delta,nowR+=delta;
if(nowR-nowL+1>mod/4)
nowL=0,nowR=mod-1;
for(int i=1;i<=S2;i++){
long long nowA=0,nowC=0;
for(int j=1;j<=S3;j++){
int x=rnd()%n+1;
while(1){
int flg=0;
for(int k=1;k<j;k++)
if(p[k]==x)
flg=1;
if(flg==0)
break;
x=rnd()%n+1;
}
p[j]=x;
long long aa=a[x],cc=inc(c[x]+mul(a[x],delta));
if(rnd()&1)
nowA=inc(nowA+aa),nowC=inc(nowC+cc);
else nowA=dec(nowA-aa),nowC=dec(nowC-cc);
}
addnode(nowA,nowC,(long long)S3);
}
for(int i=1;i<=S4;i++){
if(V.size()>S5)
nth_element(V.begin(),V.begin()+S5,V.end()),V.resize(S5);
sort(V.begin(),V.end()),V.erase(unique(V.begin(),V.end()),V.end());
int rec=V.size();
for(int j=0;j<rec;j++)
for(int k=j+1;k<=j+S6&&k<rec;k++)
addnode(V[k].a-V[j].a,dec(V[k].c-V[j].c),V[j].cnt+V[k].cnt);
}
nowL-=delta,nowR-=delta;
}
ans=0;
for(long long i=nowL;i<=nowR;i++)
if(check(i)){
ans=i;
break;
}
printf("%lld\n",(ans%mod+mod)%mod);
}
return 0;
}
詳細信息
Test #1:
score: 0
Time Limit Exceeded
input:
1 50 922033901407246477 492300877907148697 8585039545574817 36478175140515505 237143454432095134 537753813197233578 694568987600933631 82789600405017625 562554784008054548 282428338659751664 111810756866364131 822189940492446170 74373256662670811 772047021157436862 845822103589054835 905076737759700...