poj3252 数位dp
这题不是用10进制储存的，要转化成2进制再计算
dp[i][j][k] i是位数，j是1的个数，k是0的个数
#include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<cassert> #include<iomanip> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define C 0.5772156649 #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const double g=10.0,eps=1e-7; const int N=50+10,maxn=1000000+10,inf=0x3f3f3f; int dp[N][N][N],digit[N]; int dfs(int len,int one,int zero,bool fi,bool fp) { if(!len) { if(fi)return 1; return zero>=one; } if(!fp&&!fi&&dp[len][one][zero]!=-1)return dp[len][one][zero]; int ans=0,fpmax=fp ? digit[len] : 1; for(int i=0;i<=fpmax;i++) { if(fi) { if(i==0)ans+=dfs(len-1,0,0,fi,fp&&i==fpmax); else ans+=dfs(len-1,one+1,zero,fi&0,fp&&i==fpmax); } else { if(i==0)ans+=dfs(len-1,one,zero+1,fi&0,fp&&i==fpmax); else ans+=dfs(len-1,one+1,zero,fi&0,fp&&i==fpmax); } } if(!fp&&!fi)dp[len][one][zero]=ans; return ans; } ll solve(ll x) { int len=0; while(x) { digit[++len]=x&1; x/=2; } return dfs(len,0,0,1,1); } int main() { ios::sync_with_stdio(false); cin.tie(0); int a,b; memset(dp,-1,sizeof dp); cin>>a>>b; cout<<solve(b)-solve(a-1)<<endl; return 0; } /******************** ********************/
#include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<cassert> #include<iomanip> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define C 0.5772156649 #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const double g=10.0,eps=1e-7; const int N=50+10,maxn=1000000+10,inf=0x3f3f3f; int dp[N][N][N],digit[N]; int dfs(int len,int one,int zero,bool fi,bool fp) { if(!len) { return !fi&&zero>=one; } if(!fp&&!fi&&dp[len][one][zero]!=-1)return dp[len][one][zero]; int ans=0,fpmax=fp ? digit[len] : 1; for(int i=0;i<=fpmax;i++) { if(fi) { if(i==0)ans+=dfs(len-1,0,0,fi,fp&&i==fpmax); else ans+=dfs(len-1,one+1,zero,fi&0,fp&&i==fpmax); } else { if(i==0)ans+=dfs(len-1,one,zero+1,fi&0,fp&&i==fpmax); else ans+=dfs(len-1,one+1,zero,fi&0,fp&&i==fpmax); } } if(!fp&&!fi)dp[len][one][zero]=ans; return ans; } ll solve(ll x) { int len=0; while(x) { digit[++len]=x&1; x/=2; } return dfs(len,0,0,1,1); } int main() { ios::sync_with_stdio(false); cin.tie(0); int a,b; memset(dp,-1,sizeof dp); cin>>a>>b; cout<<solve(b)-solve(a-1)<<endl; return 0; } /******************** ********************/
