Urban (city) (inclusion and exclusion)

T1 on the examination

DP was found 70 min water ah, f [i] [j] i is assigned to the current position of the program j teams

We use the prefix and statistics, will be deleted i, j reverse enumeration, you can delete one-dimensional (arrays can also scroll rolled up)

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<string>
 6 #include<algorithm>
 7 #include<vector>
 8 #include<map>
 9 #define MAXN 110001
10 #define int long long
11 using namespace std;
12 int f[MAXN],sum[MAXN];
13 int n,m,K;
14 const int mod=998244353;
15 int ans=0;
16 signed main()
17 {
18   // freopen("text.in","r",stdin);
19   // freopen("a.out","w",stdout);  
20     scanf("%lld%lld%lld",&n,&m,&K);    
21      m-=n;K-=1;
22     if(m<0||K<0)
23     {
24          printf("0\n");
25          return 0;
26     }
27     f[0]=1;sum[0]=1;
28     for(int j=1;j<=m;++j)sum[j]=sum[j-1]+f[j];
29     for(int i=1;i<=n;++i)
30     {
31         for(int j=m;j>=0;--j)
32         {
33             if(j-K-1>=0)
34                f[j]=(sum[j]-sum[j-K-1]+mod)%mod;
35             else
36                f[j]=sum[j]%mod;
37            // printf("%lld f[%lld]=%lld\n",i,j,f[j]);
38         }
39         sum[0]=f[0];
40         for(int j=1;j<=m;++j)
41         {
42             sum[j]=(sum[j-1]+f[j]+mod)%mod;
43         }
44     }
45     printf("%lld\n",f[m]%mod);
46 }
47 /*
48 3 7 3
49 */

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Correct answer is the inclusion-exclusion

Every enumeration has at least i do not meet the conditions, it is clear that meet the inclusion and exclusion Theorem

And C (n, i) n represents any of cities in the selected i-th non-compliance

C (m-1-i * K, n-1) is clearly baffle method, i * K now meet the conditions of this section are not deleted, then in the remaining zone

Insert the n-1 board, it means is divided into n parts,

And because C (n, i) is selected i do not meet, and then multiplied, is at least the case when all i

We found that at least include the case of i +, i + 2 ..... case i. 1,

So we can see that the situation i + 1 can be seen as at least as when i, determine the i, they chose a

At least one example,

Selected for the case 2 is C (2,1) times (two numbers set 2 are selected from a set of repeated too)

Then add about inclusion and exclusion at least changed to 2:00

As proved more rigorous inclusion and exclusion, you can see other people blog, I will not permit the .......

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<string>
 6 #include<algorithm>
 7 #include<vector>
 8 #include<map>
 9 #define MAXN 31000001
10 #define int long long
11 using namespace std;
12 const int mod=998244353;
13 int n,m,K;
14 int ni_c[MAXN],jie[MAXN],ni[MAXN];
15 int C(int x,int y)
16 {
17     if(y>x)return 0;
18     if(y==0)return 1;
19     return jie[x]*ni_c[y]%mod*ni_c[x-y]%mod;
20 }
21 signed main()
22 {
23    scanf("%lld%lld%lld",&n,&m,&K);
24    int ans=0;
25    jie[1]=1;ni[1]=1 ; ni_c [ 1 ] = 1 ;
26     Jie [ 0 ] = 1 ; nine [ 0 ] = 1 ; ni_c [ 0 ] = 1 ;
27     for ( int i = 2 ; i <= m + 1 ; ++ i)
 28     {
 29         Jie [i] = Jie [i- 1 ] * as% v;
30         nine [i] = (v-v / i) * Nine [v% i]% v;
31         ni_c [i] = ni_c [i- 1 ] * Nine [i]% v;
32     }
 33     ans = C (m-1,n-1)%mod;
34    int base=n;
35    for(int i=1;i<=n;++i)
36    {
37        if(m-(i*K)<n)break;
38        if((i%2)==0)
39        {
40            ans=(ans+base*C(m-1-(i*K),n-1)%mod+mod)%mod;
41        }
42        else
43        {
44            ans = (ANS base * C (M 1 - (I * K), n 1 )% v + v)% v;
45         }
 46         Base = ( Base * (n) v%) * Nine [i + 1 ]% v;
47     }
 48     printf ( " % lld \ n " , ans% MOD);
49 }

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