poj3252 Round Numbers (数位dp）
Description
The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names) in order to make arbitrary decisions such as who gets to be milked first. They can't even flip a coin because it's so hard to toss using hooves.
They have thus resorted to "round number" matching. The first cow picks an integer less than two billion. The second cow does the same. If the numbers are both "round numbers", the first cow wins,
otherwise the second cow wins.
A positive integer N is said to be a "round number" if the binary representation of N has as many or more zeroes than it has ones. For example, the integer 9, when written in binary form, is 1001. 1001 has two zeroes and two ones; thus, 9 is a round number. The integer 26 is 11010 in binary; since it has two zeroes and three ones, it is not a round number.
Obviously, it takes cows a while to convert numbers to binary, so the winner takes a while to determine. Bessie wants to cheat and thinks she can do that if she knows how many "round numbers" are in a given range.
Help her by writing a program that tells how many round numbers appear in the inclusive range given by the input (1 ≤ Start < Finish ≤ 2,000,000,000).
Input
Output
Sample Input
2 12
Sample Output
6
题意：一个数位round数，要满足这个数化成二进制后0的个数大于等于1的个数。
思路：先预处理出dp[i][j]表示前i位有j个0的方案数，然后从高位到低位数位dp就行了。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 805
ll dp[40][40];
void init()
{
int i,j;
memset(dp,0,sizeof(dp));
dp[1][1]=1;dp[1][0]=1;
for(i=1;i<=32;i++){
for(j=0;j<=i;j++){
dp[i+1][j]+=dp[i][j];
dp[i+1][j+1]+=dp[i][j];
}
}
}
ll solve(int x)
{
int i,j,t=x;
int wei[40],len=0;
while(t){
wei[++len]=t%2;
t/=2;
}
ll sum=0;
int num0=0,num1=0;
for(i=len-1;i>=1;i--){ //这里先把第len位变为0，然后一次枚举最高的位数在第i位
for(j=0;j<=i-1;j++){
if(j>=i-j)sum+=dp[i-1][j];
}
}
num1=1;
for(i=len-1;i>=1;i--){ //这里是在第len位为1的情况下进行dp
if(wei[i]==1){
if(i==1){
if(num0+1>=num1)sum++;
}
else{
for(j=0;j<=i-1;j++){
if(j+num0+1>=num1+i-1-j ){
sum+=dp[i-1][j];
}
}
}
num1++;
}
else num0++;
}
return sum;
}
int main()
{
int n,m,i,j;
init();
while(scanf("%d%d",&m,&n)!=EOF)
{
printf("%lld\n",solve(n+1)-solve(m));
}
return 0;
}
这题也可以用记忆化搜索做，搜索比直接dp要好些，而且用途广,写的时候在dfs中加一维zero，表示当前这一位是不是任然是前导0.用dp[pos][num0][num1]表示前pos位0的个数为num0,1的个数为num1方案数。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
int dp[35][35][35];
int wei[35];
int dfs(int pos,int num0,int num1,int lim,int zero)
{
int i,j;
if(pos==0){
if(zero==0){ //这里不判断zero==0也可以，判不判断的区别在于是不是把0算上，判断就不把0算上了
if(num0>=num1)return 1;
else return 0;
}
return 0;
/*
if(num0>=num1)return 1;
return 0;
*/
}
if(lim==0 && dp[pos][num0][num1]!=-1){
return dp[pos][num0][num1];
}
int ed=lim?wei[pos]:1;
int ans=0,nu0,nu1;
for(i=0;i<=ed;i++){
if(zero && i==0){
nu0=nu1=0;
}
else{
if(i==0){
nu0=num0+1;
nu1=num1;
}
else{
nu0=num0;
nu1=num1+1;
}
}
ans+=dfs(pos-1,nu0,nu1,lim&&i==ed,zero&&i==0 );
}
if(lim==0){
dp[pos][num0][num1]=ans;
}
return ans;
}
int solve(int x)
{
int i,j,tot=0;
while(x){
wei[++tot]=x%2;
x/=2;
}
return dfs(tot,0,0,1,1);
}
int main()
{
int n,m,i,j;
memset(dp,-1,sizeof(dp));
while(scanf("%d%d",&m,&n)!=EOF)
{
printf("%d\n",solve(n)-solve(m-1));
}
return 0;
}