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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#131303#6137. Sub-cycle Graphstefanbalaz2WA 266ms4836kbC++141.8kb2023-07-26 21:23:132023-07-26 21:23:14

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-07-26 21:23:14]
  • 评测
  • 测评结果:WA
  • 用时:266ms
  • 内存:4836kb
  • [2023-07-26 21:23:13]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
#define ff first
#define ss second
#define pb push_back
#define ll long long
typedef pair<int,int> pii;

const int mod=1e9+7;
inline int add(int x,int y){int ret=x+y;if(ret>=mod)ret-=mod;return ret;}
inline int sub(int x,int y){int ret=x-y;if(ret<0)ret+=mod;return ret;}
inline int mul(int x,int y){return ((ll)x*y)%mod;}
inline int step(int base,int pw){int ret=1;while(pw){if(pw&1)ret=mul(ret,base);base=mul(base,base);pw>>=1;}return ret;}
inline int invv(int x){return step(x,mod-2);}

const int maxn=1e5+10;

int fact[maxn],inv[maxn],prek2[maxn];
int c(int n,int k){
    if(n<0 || k<0 || n<k)return 0;
    return mul(fact[n],mul(inv[k],inv[n-k]));
}
int pw2(int n){

    if(n<0)return 0;
    if(n==0)return 1;
    if(prek2[n]!=-1)return prek2[n];

    prek2[n]=mul(pw2(n-1),2);
    return prek2[n];
}

int main() {


    ///freopen("test.txt","r",stdin);

    inv[0]=fact[0]=1;
    for(int i=1;i<maxn;i++){
        fact[i]=mul(fact[i-1],i);
        inv[i]=invv(fact[i]);
    }
    memset(prek2,-1,sizeof(prek2));

    int t;
    scanf("%d",&t);
    while(t--){

        int n;
        ll m;
        scanf("%d %lld",&n,&m);
        ll k=n-m;
        if(k<0){
            printf("0\n");
            continue;
        }
        if(k==0){
            printf("1\n");
            continue;
        }



        int rez=0;
        for(int i=0;i<=n-1;i++){

            int oc=n-1-i;
            int pom=1;
            if(oc%2)pom=mod-1;

            rez=add( rez, mul( c(i,k-1) , mul(c(k,oc),mul(pom,pw2(k-oc)) ) ) );

        }

        rez=mul(rez,fact[n]);
        rez=mul(rez,inv[k]);
        rez=mul(rez,invv(step(2,k)));

        printf("%d\n",rez);

    }


    return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 12ms
memory: 4836kb

input:

3
4 2
4 3
5 3

output:

15
12
90

result:

ok 3 number(s): "15 12 90"

Test #2:

score: -100
Wrong Answer
time: 266ms
memory: 4784kb

input:

17446
3 0
3 1
3 2
3 3
4 0
4 1
4 2
4 3
4 4
5 0
5 1
5 2
5 3
5 4
5 5
6 0
6 1
6 2
6 3
6 4
6 5
6 6
7 0
7 1
7 2
7 3
7 4
7 5
7 6
7 7
8 0
8 1
8 2
8 3
8 4
8 5
8 6
8 7
8 8
9 0
9 1
9 2
9 3
9 4
9 5
9 6
9 7
9 8
9 9
10 0
10 1
10 2
10 3
10 4
10 5
10 6
10 7
10 8
10 9
10 10
11 0
11 1
11 2
11 3
11 4
11 5
11 6
11 7
11...

output:

1
3
3
1
1
6
15
12
1
1
10
45
90
60
1
1
15
105
375
630
360
1
1
21
210
1155
3465
5040
2520
1
1
28
378
2940
13545
35280
45360
20160
1
1
36
630
6552
42525
170100
393120
453600
181440
1
1
45
990
13230
114345
643545
2286900
4762800
4989600
1814400
1
1
55
1485
24750
273735
2047815
10239075
32848200
62370000...

result:

wrong answer 9th numbers differ - expected: '3', found: '1'