poj 3252 Round Numbers(数位dp)
2017-02-26 00:38
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题目地址:点击打开链接
题意:
Round Numbers:二进制0比1多或者相等的数。
给定一个区间,问在这个区间上的Round Numbers有多少个?
思路:裸的数位dp,做过处理前导0的数位dp就没问题。
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 35;
int dp[maxn][maxn][maxn], a[maxn];
int dfs(int pos, int num0, int num1, int lead, int limit)
{
if(pos == -1) return num0 >= num1;
if(!limit && lead && dp[pos][num0][num1] != -1) return dp[pos][num0][num1];
int up = limit ? a[pos] : 1;
int tmp = 0;
for(int i = 0; i <= up; i++)
{
if(lead)
tmp += dfs(pos-1, num0+(i==0), num1+(i==1), lead || i, limit && i==a[pos]);
else
tmp += dfs(pos-1, 0, num1+(i==1), lead || i, limit && i==a[pos]);
}
if(!limit && lead) dp[pos][num0][num1] = tmp;
return tmp;
}
int solve(int x)
{
int pos = 0;
while(x)
{
a[pos++] = x%2;
x /= 2;
}
return dfs(pos-1, 0, 0, 0, 1);
}
int main(void)
{
int l, r;
memset(dp, -1, sizeof(dp));
while(cin >> l >> r)
printf("%d\n", solve(r)-solve(l-1));
return 0;
}
Round Numbers
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
Line 1: Two space-separated integers, respectively Start and Finish.
Output
Line 1: A single integer that is the count of round numbers in the inclusive range Start..Finish
Sample Input
Sample Output
Source
USACO 2006 November Silver
题意:
Round Numbers:二进制0比1多或者相等的数。
给定一个区间,问在这个区间上的Round Numbers有多少个?
思路:裸的数位dp,做过处理前导0的数位dp就没问题。
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 35;
int dp[maxn][maxn][maxn], a[maxn];
int dfs(int pos, int num0, int num1, int lead, int limit)
{
if(pos == -1) return num0 >= num1;
if(!limit && lead && dp[pos][num0][num1] != -1) return dp[pos][num0][num1];
int up = limit ? a[pos] : 1;
int tmp = 0;
for(int i = 0; i <= up; i++)
{
if(lead)
tmp += dfs(pos-1, num0+(i==0), num1+(i==1), lead || i, limit && i==a[pos]);
else
tmp += dfs(pos-1, 0, num1+(i==1), lead || i, limit && i==a[pos]);
}
if(!limit && lead) dp[pos][num0][num1] = tmp;
return tmp;
}
int solve(int x)
{
int pos = 0;
while(x)
{
a[pos++] = x%2;
x /= 2;
}
return dfs(pos-1, 0, 0, 0, 1);
}
int main(void)
{
int l, r;
memset(dp, -1, sizeof(dp));
while(cin >> l >> r)
printf("%d\n", solve(r)-solve(l-1));
return 0;
}
Round Numbers
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 12849 | Accepted: 4961 |
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
Line 1: Two space-separated integers, respectively Start and Finish.
Output
Line 1: A single integer that is the count of round numbers in the inclusive range Start..Finish
Sample Input
2 12
Sample Output
6
Source
USACO 2006 November Silver
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