1032 - Fast Bit Calculations
PDF (English) Statistics Forum
Time Limit: 2 second(s) Memory Limit: 32 MB
A bit is a binary digit, taking a logical value of either 1or 0 (also referred to as "true" or "false"respectively). And every decimal number has a binary representation which isactually a series of bits. If a bit of a number is 1 and its next bit isalso 1 then we can say that the number has a 1 adjacent bit. Andyou have to find out how many times this scenario occurs for all numbers up to N.
Examples:
Number Binary Adjacent Bits
12 1100 1
15 1111 3
27 11011 2
Input
Input starts with an integer T (≤ 10000),denoting the number of test cases.
Each case contains an integer N (0 ≤ N < 231).
Output
For each test case, print the case number and the summationof all adjacent bits from 0 to N.
Sample Input
Output for Sample Input
7
0
6
15
20
21
22
2147483647
Case 1: 0
Case 2: 2
Case 3: 12
Case 4: 13
Case 5: 13
Case 6: 14
Case 7: 16106127360
题意：对于一个数 n ，改成二进制的数 ， Adjacent Bits就是这个二进制数中连续的 1 的个数（含有11个数），给你 n，让你求 1—>n里面有多少个Adjacent Bits
分析：
数位DP的模板题，dp[pos][sta][pre]表示到当前位置pos，前面有多少个11，前一位是pre的，答案是多少。
如果前一位是1，后以为是1，记录下来，否则向下枚举即可。
///求前n个数的二进制形式含有11个数的和
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=40;
typedef long long ll;
int n;
int dig[maxn];
ll dp[maxn][maxn][2];
///dp[pos][sta][pre]表示到当前位置pos，前面有多少个11，前一位是pre的，答案是多少
ll dfs(int pos,int sta,int limit,int pre)
{
if(pos<0) return sta;
if(!limit&&dp[pos][sta][pre]!=-1)
return dp[pos][sta][pre];
ll ans=0;
int up=(limit?dig[pos]:1); ///二进制的时候为1，十进制的时候9
for(int i=0;i<=up;i++)
{
if(pre&&i)///前一个是1，后一个也是1。记录个数
ans+=dfs(pos-1,sta+1,limit&&i==up,i);
else
ans+=dfs(pos-1,sta,limit&&i==up,i);
}
if(!limit)dp[pos][sta][pre]=ans;
return ans;
}
ll solve(int n)
{
int len=0;
memset(dp,-1,sizeof(dp));
while(n)
{
dig[len++]=n%2; ///转化为二进制
n/=2;
}
return dfs(len-1,0,1,0);
}
int main()
{
int T,cas=1;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
printf("Case %d: %lld\n",cas++,solve(n));
}
return 0;
}
