QOJ.ac

QOJ

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
#823817#8428. Partition into TeamsInvincibleWA 1ms6000kbC++231.8kb2024-12-21 09:58:302024-12-21 09:58:30

Judging History

你现在查看的是最新测评结果

  • [2024-12-21 09:58:30]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:6000kb
  • [2024-12-21 09:58:30]
  • 提交

answer

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <ctime>
#include <random>
#include <cassert>
#include <numeric>
#include <cmath>
#include <bitset>
#include <ext/pb_ds/assoc_container.hpp>
#define pii pair<int, int>
#define fi first
#define se second
#define MP make_pair
#define ep emplace
#define eb emplace_back
#define int long long
#define rep(i, j, k) for (int i = (j); i <= (k); i++)
#define per(i, j, k) for (int i = (j); i >= (k); i--)
typedef double db;
typedef long double ldb;
typedef long long ll;
//typedef __int128 lll;
typedef unsigned long long ull;
typedef unsigned int ui;
using namespace std;
using namespace __gnu_pbds;
bool Mbe;

//char buf[1<<20],*p1,*p2;
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin), p1 == p2) ? 0 : *p1++)
int read() {
	int s = 0, f = 1;
	char c = getchar();
	while (c < '0' || c > '9') f ^= (c == '-'), c = getchar();
	while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();
	return f ? s : -s;
}
template<typename T>void chkmax(T&x,const T&y){if(x<y)x=y;}
template<typename T>void chkmin(T&x,const T&y){if(x>y)x=y;}

const int N=1000005;
int n,p,fac[N],ifac[N],ans;
int fpow(int x,int k=p-2){
	int res=1;
	for(;k;k>>=1){
		if(k&1)res=res*x%p;
		x=x*x%p;
	}
	return res;
}
int C(int x,int y){return x<y?0:fac[x]*ifac[y]%p*ifac[x-y]%p;}
int Comb(int x,int y){
	if(x<y)return 0;
	if(x<p&&y<p)return C(x,y);
	return C(x%p,y%p)*Comb(x/p,y/p)%p;
}

bool Med;
signed main() {
	fprintf(stderr,"%.3lfMb\n",(&Mbe-&Med)/1024./1024.);
	n=read(),p=read();
	fac[0]=1;
	rep(i,1,p-1)fac[i]=fac[i-1]*i%p;
	ifac[p-1]=fpow(fac[p-1]);
	per(i,p-1,1)ifac[i-1]=ifac[i]*i%p;
	rep(i,0,min(n,p))ans=(ans+Comb(n,i)*Comb(n-i,i))%p;
	ans=(fpow(3,n)-ans+p)*fpow(2)%p;
	printf("%lld\n",ans);
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 5920kb

input:

5 5

output:

1

result:

ok 1 number(s): "1"

Test #2:

score: 0
Accepted
time: 1ms
memory: 6000kb

input:

5 7

output:

5

result:

ok 1 number(s): "5"

Test #3:

score: -100
Wrong Answer
time: 1ms
memory: 5872kb

input:

789 97

output:

47

result:

wrong answer 1st numbers differ - expected: '53', found: '47'