【BZOJ】【1662】/【POJ】【3252】 【USACO 2006 Nov】Round Number
数位DP
同上一题Windy数
预处理求个组合数
然后同样的方法,这次是记录一下0和1的个数然后搞搞
Orz cxlove
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 /************************************************************** 2 Problem: 1662 3 User: Tunix 4 Language: C++ 5 Result: Accepted 6 Time:0 ms 7 Memory:1280 kb 8 ****************************************************************/ 9 10 //POJ 3252 && BZOJ 1662 11 #include<cmath> 12 #include<vector> 13 #include<cstdio> 14 #include<cstring> 15 #include<cstdlib> 16 #include<iostream> 17 #include<algorithm> 18 #define rep(i,n) for(int i=0;i<n;++i) 19 #define F(i,j,n) for(int i=j;i<=n;++i) 20 #define D(i,j,n) for(int i=j;i>=n;--i) 21 #define pb push_back 22 using namespace std; 23 int getint(){ 24 int v=0,sign=1; char ch=getchar(); 25 while(!isdigit(ch)) {if(ch=='-') sign=-1; ch=getchar();} 26 while(isdigit(ch)) {v=v*10+ch-'0'; ch=getchar();} 27 return v*sign; 28 } 29 const int N=1e7+10,INF=~0u>>2; 30 const double eps=1e-8; 31 /*******************template********************/ 32 int c[40][40]; 33 void init(){ 34 c[0][0]=1; 35 F(i,1,32){ 36 c[i][0]=c[i-1][0]; 37 F(j,1,i) c[i][j]=c[i-1][j]+c[i-1][j-1]; 38 } 39 } 40 int solve(int n){ 41 int len=0,bit[40],cnt=0; 42 for(;n;n>>=1) bit[++len]=n&1; 43 int ans=0; 44 F(i,1,len-1) 45 D(j,i-1,(i+1)>>1) 46 ans+=c[i-1][j]; 47 int one=1,zero=0; 48 D(i,len-1,1) 49 if (bit[i]){ 50 zero++; 51 F(j,max(0,(len+1)/2-zero),i-1) 52 ans+=c[i-1][j]; 53 zero--; 54 one++; 55 } 56 else zero++; 57 // ans+=c[cnt-1][(len+1)>>1]; 58 return ans; 59 } 60 int main(){ 61 init(); 62 int l=getint(),r=getint(); 63 printf("%d\n",solve(r+1)-solve(l)); 64 return 0; 65 }