HDU 5136 Yue Fei's Battle
题目链接：HDU-5136
网上的一篇题解非常好，所以就直接转载了。转自oilover的博客
代码：
1 #include<cstring> 2 #include<cstdio> 3 #include<queue> 4 using namespace std; 5 typedef long long LL; 6 const LL MAXN=100000; 7 const LL MOD=1000000007; 8 9 LL dp[MAXN+10],sum[MAXN+10]; 10 LL mod2,mod6; 11 LL extgcd(LL a,LL b,LL &x,LL &y) 12 { 13 LL d=a; 14 if(b!=0) 15 { 16 d=extgcd(b,a%b,y,x); 17 y-=(a/b)*x; 18 } 19 else { x=1; y=0; } 20 return d; 21 } 22 LL modInverse(LL a,LL m) 23 { 24 LL x,y; 25 extgcd(a,m,x,y); 26 return (m+x%m)%m; 27 } 28 void init() 29 { 30 mod2=modInverse(2,MOD); 31 mod6=modInverse(6,MOD); 32 memset(dp,0,sizeof(dp)); 33 memset(sum,0,sizeof(sum)); 34 dp[0]=dp[1]=1; 35 sum[0]=1; 36 sum[1]=2; 37 for(LL i=2;i<=MAXN;i++) 38 { 39 dp[i]=(dp[i-1]*sum[i-2] % MOD + dp[i-1]*(dp[i-1]+1) %MOD *mod2 % MOD)%MOD; 40 sum[i]=(sum[i-1]+dp[i])%MOD; 41 } 42 } 43 LL f(LL kk) 44 { 45 LL k=kk/2; 46 if(kk%2==0) 47 return dp[k]*(dp[k]+1)%MOD *mod2 % MOD; 48 LL ans=0; 49 ans =(ans + dp[k]*(dp[k]+1)%MOD *mod2 %MOD *sum[k-1] %MOD) %MOD; 50 ans =(ans + dp[k]) %MOD; 51 ans =(ans + dp[k]*((dp[k]-1+MOD)%MOD) %MOD) %MOD; 52 if(dp[k]>=3) ans =(ans + dp[k]*(dp[k]-1)%MOD*(dp[k]-2)%MOD*mod6%MOD)%MOD; 53 return ans; 54 } 55 int main() 56 { 57 #ifdef LOCAL 58 freopen("in.txt","r",stdin); 59 // freopen("out.txt","w",stdout); 60 #endif 61 init(); 62 LL k; 63 while(scanf("%lld",&k)!=EOF && k) 64 printf("%lld\n",f(k)); 65 return 0; 66 }
