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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#515130 | #4883. Bayan Testing | uuku | WA | 4ms | 7700kb | C++14 | 4.1kb | 2024-08-11 15:26:34 | 2024-08-11 15:26:34 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
namespace ModInt{
template <uint32_t mod>
struct mint
{
#define i32 int32_t
#define u32 uint32_t
#define u64 uint64_t
static constexpr u32 get_r(){u32 ret=mod;for(i32 i=0;i<4;++i)ret*=2-mod*ret;return ret;}
static constexpr u32 r=get_r();
static const u32 n2=-u64(mod)%mod;
static const u32 mod2=mod<<1;
u32 a;
constexpr mint():a(0){}
constexpr mint(const int64_t &b):a(reduce(u64(b%mod+mod)*n2)){};
static constexpr u32 reduce(const u64 &b){return (b+u64(u32(b)*u32(-r))*mod)>>32;}
const mint &operator+=(const mint &b){if(i32(a+=b.a-mod2)<0)a+=mod2;return *this;}
const mint &operator-=(const mint &b){if(i32(a-=b.a)<0)a+=mod2;return *this;}
const mint &operator*=(const mint &b){a=reduce(u64(a)*b.a);return *this;}
const mint &operator/=(const mint &b){*this*=b.inverse();return *this;}
const mint operator+(const mint &b)const{return mint(*this)+=b;}
const mint operator-(const mint &b)const{return mint(*this)-=b;}
const mint operator*(const mint &b)const{return mint(*this)*=b;}
const mint operator/(const mint &b)const{return mint(*this)/=b;}
const bool operator==(const mint &b)const{return(a>=mod?a-mod:a)==(b.a>=mod?b.a-mod:b.a);}
const bool operator!=(const mint &b)const{return(a>=mod?a-mod:a)!=(b.a>=mod?b.a-mod:b.a);}
const mint operator-()const{return mint()-mint(*this);}
const mint ksm(u64 n)const{mint ret(1);for(mint mul(*this);n;n>>=1,mul*=mul)if(n&1)ret*=mul;return ret;}
const mint inverse()const{return ksm(mod-2);}
friend ostream &operator<<(ostream &os, const mint &b){return os<<b.get();}
friend istream &operator>>(istream &is, mint &b){int64_t t;is>>t;b=mint(t);return(is);}
const u32 get()const{u32 ret=reduce(a);return ret>=mod?ret-mod:ret;}
static const u32 get_mod(){return mod;}
};
}
using namespace ModInt;
namespace FAST_IO{
#define ll long long
#define ull unsigned long long
#define db double
#define _8 __int128_t
#define Get() (BUF[Pin++])
const int LEN=1<<20;
char BUF[LEN];
int Pin=LEN;
inline void flushin(){memcpy(BUF,BUF+Pin,LEN-Pin),fread(BUF+LEN-Pin,1,Pin,stdin),Pin=0;return;}
inline char Getc(){return (Pin==LEN?(fread(BUF,1,LEN,stdin),Pin=0):0),BUF[Pin++];}
template<typename tp>inline tp read(){(Pin+40>=LEN)?flushin():void();tp res=0;char f=1,ch=' ';for(;ch<'0'||ch>'9';ch=Get())if(ch=='-')f=-1;for(;ch>='0'&&ch<='9';ch=Get())res=(res<<3)+(res<<1)+ch-48;return res*f;}
template<typename tp>inline void read(tp &n){(Pin+40>=LEN)?flushin():void();tp res=0;char f=1,ch=' ';for(;ch<'0'||ch>'9';ch=Get())if(ch=='-')f=-1;for(;ch>='0'&&ch<='9';ch=Get())res=(res<<3)+(res<<1)+ch-48;n=res*f;return;}
inline int readstr(char *s){int len=0;char ch=Getc();while(!isalnum(ch))ch=Getc();while(isalnum(ch))s[len++]=ch,ch=Getc();return len;}
#define Put(x) (PUF[Pout++]=x)
char PUF[LEN];
int Pout;
inline void flushout(){fwrite(PUF,1,Pout,stdout),Pout=0;return;}
inline void Putc(char x){if(Pout==LEN)flushout(),Pout=0;PUF[Pout++]=x;}
template<typename tp>inline void write(tp a,char b='\n'){static int stk[40],top;(Pout+50>=LEN)?flushout():void();if(a<0)Put('-'),a=-a;else if(a==0)Put('0');for(top=0;a;a/=10)stk[++top]=a%10;for(;top;--top)Put(stk[top]^48);Put(b);return;}
inline void wt_str(string s){for(char i:s)Putc(i);return;}
}
using namespace FAST_IO;
#define pii pair<int,int>
#define fi first
#define se second
#define ls (rt<<1)
#define rs (rt<<1|1)
#define Ls (tr[rt].lc)
#define Rs (tr[rt].rc)
const int N=2e5+10;
int T,n,m,ans[N],tot;
struct Seg{
int l,r;
}a[N];
bool cmp(Seg a,Seg b)
{
return a.r==b.r?a.l>b.l:a.r<b.r;
}
int main()
{
read(T);
while(T--)
{
read(n),read(m);
tot=0;
for(int i=1,l,r;i<=2*m;i++)
{
read(l),read(r);
if(l!=r)a[++tot]={l,r};
}
if(tot<m)
{
write(-1);
continue;
}
sort(a+1,a+tot+1,cmp);
for(int i=1;i<=a[tot-m].r;i++)ans[i]=i;
if(m)ans[a[tot-m+1].l]=a[tot-m].r+1;
for(int i=a[tot-m].r+1;i<=n;i++)ans[i]=a[tot-m].r+1;
for(int i=1;i<=n;i++)
write(ans[i],i<n?' ':'\n');
}
flushout();
return 0;
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 7700kb
input:
3 2 1 1 1 2 2 6 2 1 3 4 6 2 4 3 5 4 3 1 2 1 1 2 2 2 3 3 3 3 4
output:
-1 1 2 5 4 5 5 1 1 1 1
result:
ok ok (3 test cases)
Test #2:
score: 0
Accepted
time: 4ms
memory: 6280kb
input:
100000 2 1 1 1 2 2 2 1 2 2 1 2 2 1 1 2 2 2 2 1 1 2 2 2 2 1 2 2 1 2 2 1 2 2 1 1 2 1 1 1 2 2 2 1 1 2 2 2 2 1 2 2 1 1 2 1 2 2 1 2 2 1 2 2 1 1 2 1 2 2 1 2 2 1 2 2 1 2 2 1 2 2 1 1 2 1 2 2 1 1 2 1 1 2 2 2 2 1 2 2 1 2 2 1 2 2 1 2 2 1 2 2 1 2 2 1 1 1 2 2 2 1 2 2 1 2 2 1 1 1 2 2 2 1 1 1 2 2 2 1 2 2 1 2 2 1 2...
output:
-1 1 1 1 1 1 1 1 1 -1 -1 1 1 -1 1 1 -1 1 1 1 1 -1 -1 1 1 1 1 1 1 1 1 -1 1 1 -1 -1 1 1 -1 1 1 1 1 -1 -1 1 1 1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 1 1 1 1 1 1 -1 1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 1 1 1 1 -1 -1 1 1 -1 1 1 1 1 -1 1 1 -1 1 1 1 1 -1 1 1 1 1 1 1 -...
result:
ok ok (100000 test cases)
Test #3:
score: -100
Wrong Answer
time: 3ms
memory: 6596kb
input:
25000 10 5 4 10 1 4 9 9 2 9 5 8 1 8 1 5 5 5 4 9 6 6 11 5 9 11 4 7 2 2 2 5 8 10 3 11 2 4 4 8 3 10 4 6 5 2 1 3 4 4 1 5 5 5 6 3 4 6 3 3 1 5 3 6 1 1 1 3 7 3 4 4 3 5 1 6 3 4 2 3 1 2 7 3 3 4 1 5 6 7 2 6 3 5 2 3 5 2 5 5 4 5 2 3 1 1 10 5 3 6 4 5 3 3 6 9 2 5 9 10 5 6 5 7 1 4 8 9 11 5 1 10 2 11 6 9 2 6 6 6 8 ...
output:
1 2 3 4 6 6 6 6 6 6 1 2 3 8 5 6 7 8 8 8 8 1 1 1 1 1 4 2 3 4 4 4 1 2 4 4 4 4 4 6 2 3 4 5 6 6 1 1 1 1 1 1 2 7 4 5 6 7 7 7 7 10 2 3 4 5 6 7 8 9 10 10 1 7 3 4 5 6 7 7 1 2 3 8 5 6 7 8 8 8 1 6 3 4 5 6 6 6 1 2 10 4 5 6 7 8 9 10 10 10 10 10 10 10 10 1 2 3 4 5 6 7 10 9 10 10 10 10 3 2 3 3 3 3 1 2 8 4 5 6 7 8...
result:
wrong answer the number of segments with two equal elements is 1 != 3 (test case 6)