A - B = C 可以转化成 B + C = A
题解：http://m.blog.csdn.net/blog/stl112514/48678307
http://acm.hdu.edu.cn/showproblem.php?pid=5456
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; typedef long long ll; int n,m; int a[10]={6,2,5,5,4,5,6,3,7,6}; ll dp[510][2][2][2]; int dfs(int num,int cr,int b,int c){ if(num>n) return 0; if(dp[num][cr][b][c]!=-1) return dp[num][cr][b][c]; if(b==1&&c==1){ //两个加数都已经到最高位 if(num==n&&cr==0) //没有进位 return 1; if(num+a[1]==n&&cr==1) //有进位，剩余的火柴数刚好等于进位"1" return 1; return 0; } if(num==n) //火柴已经用完，又不符合上面的判别条件 return 0; dp[num][cr][b][c]=0; if(b==0){ if(c==0){ for(int i=0;i<10;i++){ for(int j=0;j<10;j++){ int t=a[i]+a[j]+a[(i+j+cr)%10]; dp[num][cr][b][c]+=dfs(num+t,(i+j+cr)/10,0,0); if(i!=0) dp[num][cr][b][c]+=dfs(num+t,(i+j+cr)/10,1,0); if(j!=0) dp[num][cr][b][c]+=dfs(num+t,(i+j+cr)/10,0,1); if(i!=0&&j!=0) dp[num][cr][b][c]+=dfs(num+t,(i+j+cr)/10,1,1); dp[num][cr][b][c]%=m; } } } else{ for(int i=0;i<10;i++){ int t=a[i]+a[(i+cr)%10]; dp[num][cr][b][c]+=dfs(num+t,(i+cr)/10,0,1); if(i!=0) dp[num][cr][b][c]+=dfs(num+t,(i+cr)/10,1,1); } } } else{ if(c==0){ for(int i=0;i<10;i++){ int t=a[i]+a[(i+cr)%10]; dp[num][cr][b][c]+=dfs(num+t,(i+cr)/10,1,0); if(i!=0) dp[num][cr][b][c]+=dfs(num+t,(i+cr)/10,1,1); } } } return dp[num][cr][b][c]%=m; } int main(){ int T; scanf("%d",&T); for(int cs=1;cs<=T;cs++){ cin>>n>>m; n-=3; memset(dp,-1,sizeof(dp)); printf("Case #%d: %d\n",cs,dfs(0,0,0,0)); } }
#include #include #include #include//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0}; #include//int gcd(int a,int b){return b?gcd(b,a%b):a;} #include #include #include #include #include #include #define mod 1e9+7 #define ll long long using namespace std; int n,m; ll dp[505][2][2][2]; //不知道为什么int不行。。 int x[10]={6,2,5,5,4,5,6,3,7,6}; int dfs(int num,int f,int a,int b){ if(num>n) return 0; if(dp[num][f][a][b]!=-1) return dp[num][f][a][b]; if(num==n){ if(f==0&&a==1&&b==1) return 1; return 0; } if(a==1&&b==1){ if(f==1&&num+2==n) return 1; return 0; } dp[num][f][a][b]=0; if(a==0&&b==0){ for(int i=0;i<10;++i){ for(int j=0;j<10;++j){ dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[j]+x[(i+j+f)%10],(i+j+f)/10,0,0))%m; if(i!=0) dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[j]+x[(i+j+f)%10],(i+j+f)/10,1,0))%m; if(j!=0) dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[j]+x[(i+j+f)%10],(i+j+f)/10,0,1))%m; if(i!=0&&j!=0) dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[j]+x[(i+j+f)%10],(i+j+f)/10,1,1))%m; } } } else if(a==1&&b==0){ for(int j=0;j<10;++j){ dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[j]+x[(j+f)%10],(j+f)/10,1,0))%m; if(j!=0) dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[j]+x[(j+f)%10],(j+f)/10,1,1))%m; } } else if(a==0&&b==1){ for(int i=0;i<10;++i){ dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[(i+f)%10],(i+f)/10,0,1))%m; if(i!=0) dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[(i+f)%10],(i+f)/10,1,1))%m; } } return dp[num][f][a][b]; } int main(){ int t,cnt=0; scanf("%d",&t); while(t--){ scanf("%d%d",&n,&m); n-=3; memset(dp,-1,sizeof(dp)); printf("Case #%d: %d\n",++cnt,dfs(0,0,0,0)); } return 0; }
