dp--递推

Posted 2020-10-01 yijiull

 

Attack on Titans

 ZOJ - 3747 

题意：有三种兵A，B，C，选n个排队，问（至少m个连续的B和至多连续的C）的方案数。

理解了很长时间，，，好菜=_=||

先贴上代码，有空来写

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int maxn = 1000010;
 5 const int mod = 1000000007;
 6 ll d[maxn][4];
 7 ll n,m,k;
 8 ll cal(int u){
 9     d[0][0]=1;
10     d[0][1]=0;
11     d[0][2]=0;
12     for(int i=1;i<=n;i++){
13         ll sum=(d[i-1][0]+d[i-1][1]+d[i-1][2])%mod;
14         d[i][2]=sum;
15 
16         if(i<=u) d[i][0]=sum;
17         else if(i==u+1) d[i][0]=(sum-1)%mod;
18         else d[i][0]=(sum-d[i-u-1][1]-d[i-u-1][2])%mod;
19 
20         if(i<=k) d[i][1]=sum;
21         else if(i==k+1) d[i][1]=(sum-1)%mod;
22         else d[i][1]=(sum-d[i-1-k][0]-d[i-1-k][2])%mod;
23     }
24     return (d[n][0]+d[n][1]+d[n][2])%mod;
25 }
26 int main(){
27     while(scanf("%d%d%d",&n,&m,&k)!=EOF){
28         ll ans=cal(n);
29         ans=((ans-cal(m-1))%mod+mod)%mod;
30         printf("%lld\n",ans);
31     }
32     return 0;
33 }

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