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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#267056#6137. Sub-cycle GraphzlxFTHAC ✓190ms11204kbC++172.1kb2023-11-26 21:50:582023-11-26 21:50:58

Judging History

你现在查看的是最新测评结果

  • [2023-11-26 21:50:58]
  • 评测
  • 测评结果:AC
  • 用时:190ms
  • 内存:11204kb
  • [2023-11-26 21:50:58]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define max_n 511111
#define mo 1000000007
void read(int &p)
{
    p = 0;
    int k = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
        {
            k = -1;
        }
        c = getchar();
    }
    while (c >= '0' && c <= '9')
    {
        p = p * 10 + c - '0';
        c = getchar();
    }
    p *= k;
    return;
}
void write_(int x)
{
    if (x < 0)
    {
        putchar('-');
        x = -x;
    }
    if (x > 9)
    {
        write_(x / 10);
    }
    putchar(x % 10 + '0');
}
void writesp(int x)
{
    write_(x);
    putchar(' ');
}
void writeln(int x)
{
    write_(x);
    putchar('\n');
}
int T,n,m;
int ksm(int a,int b)
{
    int pi = 1;
    for(;b;b >>= 1,a = (a * a) % mo)
    {
        if(b & 1)
        {
            pi = (pi * a) % mo;
        }
    }
    return pi;
}
int jiecheng[max_n],inv[max_n];
int ans = 0;
void solution()
{
    read(n),read(m);

    if(m > n)
    {
        puts("0");
        return ;
    }
    if(m == 0)
    {
        puts("1");
        return ;
    }
    if(n == m)
    {
        writeln((ksm(2,mo - 2) * jiecheng[n - 1] ) % mo);
        return ;
    }
    ans = 0;
    int pi = 1;
    for(int i = 1;i + m <= n && i <= m;++i)
    {
        (ans += 
        pi *
        jiecheng[m - i] % mo*
        jiecheng[n] % mo * 
        inv[n - m - i] % mo * 
        inv[m + i] % mo * 
        jiecheng[m + i] % mo * 
        inv[i << 1LL] % mo *
        inv[m - i] % mo *
        jiecheng[m - 1] % mo * 
        inv[i - 1] % mo *
        inv[m - i] % mo) %= mo; 
        pi = (pi *((i<<1) | 1) % mo);
    }
    writeln(ans);
}
signed main()
{  
    jiecheng[0] = 1;
    for(int i = 1;i <= 400000;i++)
    {
        jiecheng[i] = (jiecheng[i - 1] * i) % mo;
    }
    inv[400000] = ksm(jiecheng[400000],mo - 2);
    for(int i = 399999;~i;i--)
    {
        inv[i] = (inv[i + 1] * (i + 1)) % mo;
    }
    read(T);
    while(T--)
    {
        solution();
    }
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 11064kb

input:

3
4 2
4 3
5 3

output:

15
12
90

result:

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

Test #2:

score: 0
Accepted
time: 190ms
memory: 11204kb

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

result:

ok 17446 numbers