这题的DP很难想,定义dp[i][j][a][b]表示用了i个男生,j个女生,任一连续的后缀区间内,男生比女生最多多a人,女生比男生最多多b人。
转移就是显然了。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-9 # define MOD 12345678 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=10005; //Code begin... int dp[155][155][25][25]; int main () { int n, m, k; scanf("%d%d%d",&n,&m,&k); dp[0][0][0][0]=1; FOR(i,0,n) FOR(j,0,m) FOR(l1,0,k) FOR(l2,0,k) { dp[i+1][j][l1+1][max(l2-1,0)]=(dp[i+1][j][l1+1][max(l2-1,0)]+dp[i][j][l1][l2])%MOD; dp[i][j+1][max(l1-1,0)][l2+1]=(dp[i][j+1][max(l1-1,0)][l2+1]+dp[i][j][l1][l2])%MOD; } int ans=0; FOR(i,0,k) FOR(j,0,k) ans=(ans+dp[n][m][i][j])%MOD; printf("%d ",ans); return 0; }