UVA 10564 计数DP

也是经典的计数DP题，想练练手，故意不写记忆化搜索，改成递推，还是成功了嘞。。。不过很遗憾一开始WA了，原来是因为判断结束条件写个 n或s为0，应该要一起为0的，搞的我以为自己递推写挫了，又改了一下，其实递推没问题，就是写出来不好看

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
const int N=80;
LL dp[N][N][1000];
int mat[N][N];
int p[N][N][1000];
int n,s;
void print(int d,int loc,int s)
{
    if (d==2*n-1) return;
    int val=mat[d][loc];
    if (d<n){
      if (dp[d+1][loc-1][s-val] && loc>1){
        putchar('L');
        print(d+1,loc-1,s-val);
        return;
      }
      else {
          putchar('R');
          print(d+1,loc,s-val);
      }
    }
    else{
        if (dp[d+1][loc][s-val]){
            putchar('L');
            print(d+1,loc,s-mat[d][loc]);
            return;
        }
        else{
            putchar('R');
            print(d+1,loc+1,s-mat[d][loc]);
        }
    }
}
int main()
{
    while (scanf("%d%d",&n,&s)!=EOF)
    {
        if (n==0 &&s==0) break;
        memset(mat,0,sizeof mat);
        for (int i=1;i<=2*n-1;i++){
            int k;
            if (i<=n) k=n-i+1;
            else k=i-n+1;
            for (int j=1;j<=k;j++){
                scanf("%d",&mat[i][j]);
            }
        }
        memset(dp,0,sizeof dp);
        for (int i=1;i<=n;i++){
            dp[2*n-1][i][mat[2*n-1][i]]=1;
        }
        for (int i=2*n-2;i>=1;i--)
        {
            int k;
            if (i>=n) k=i-n+1;
            else k=n-i+1;
            if (i>=n){
                for (int j=1;j<=k;j++){
                    for (int q=mat[i][j];q<=s;q++){
                        dp[i][j][q]+=dp[i+1][j][q-mat[i][j]];
                        dp[i][j][q]+=dp[i+1][j+1][q-mat[i][j]];
                    }
                }
            }
            else{
                for (int j=1;j<=k;j++){
                    for (int q=mat[i][j];q<=s;q++){
                        dp[i][j][q]+=dp[i+1][j][q-mat[i][j]];
                        dp[i][j][q]+=dp[i+1][j-1][q-mat[i][j]];
                    }
                }
            }
        }
        LL ans=0;
        int loc=-1;
        for (int i=1;i<=n;i++){
            ans+=dp[1][i][s];
            if (dp[1][i][s] && loc==-1){
                loc=i;
            }
        }
        printf("%lld\n",ans);
        if (ans>0)
        {
            printf("%d ",loc-1);
            print(1,loc,s);
        }
        puts("");
    }
    return 0;
}