Codeforces Round #247 (Div. 2) C k-Tree

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dp[i][j][0]//第i层和为j不满足条件的个数

dp[i][j][1]//第i层和为j满足条件个数

#include<iostream>#include<sstream>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<stack>#include<math.h>#include<map>#include<time.h>#include<set>#include<string>#include<vector>#include<algorithm>using namespace std;#define inf 0x7fffffff#define lc l,m,index<<1#define rc m+1,r,index<<1|1#define max_n 200005#define mod 1000000007#define LL  __int64LL dp[105][105][2];int n,k,d;int main(){    while(~scanf("%d%d%d",&n,&k,&d))    {        memset(dp,0,sizeof(dp));        for(int i=1;i<=k;i++)        {            if(i<d)            dp[1][i][0]=1;            else            dp[1][i][1]=1;        }        for(int i=2;i<=n+1;i++)        {            for(int j=i;j<=n;j++)            {                for(int x=1;x<=k;x++)                {                    if(x>=j)continue;                    if(x<d)                    {                        dp[i][j][0]=(dp[i][j][0]+dp[i-1][j-x][0])%mod;                        dp[i][j][1]=(dp[i][j][1]+dp[i-1][j-x][1])%mod;                    }                    else                    dp[i][j][1]=(dp[i][j][1]+dp[i-1][j-x][0]+dp[i-1][j-x][1])%mod;                   // printf("i == %d j == %d x == %d  ans == %d\n",i,j,x,dp[i][j][1]);                   // system("pause");                }            }        }        LL ans=0;      /*  for(int i=1;i<=n;i++)        {            for(int j=i;j<=n;j++)            {                printf("dp[%d][%d][1] == %d\n",i,j,dp[i][j][1]);                printf("dp[%d][%d][0] == %d\n",i,j,dp[i][j][0]);            }        }        */        for(int i=1;i<=n+1;i++)        {           // printf("i == %d  ans == %d\n",i,dp[i][n][1]);            ans=(ans+dp[i][n][1])%mod;        }        printf("%I64d\n",ans);    }    return 0;}

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