[暑假集训--数位dp]LightOj1032 Fast Bit Calculations
A bit is a binary digit, taking a logical value of either 1 or 0 (also referred to as "true" or "false" respectively). And every decimal number has a binary representation which is actually a series of bits. If a bit of a number is 1 and its next bit is also 1 then we can say that the number has a 1 adjacent bit. And you 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 summation of all adjacent bits from 0 to N.
Sample Input
7
0
6
15
20
21
22
2147483647
Sample Output
Case 1: 0
Case 2: 2
Case 3: 12
Case 4: 13
Case 5: 13
Case 6: 14
Case 7: 16106127360
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define mkp(a,b) make_pair(a,b)
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
LL x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
LL n,len,l,r;
LL f[][][];
int d[];
int zhan[],top;
inline LL dfs(int now,int dat,int tot,int fp)
{
if (now==)return tot;
if (!fp&&f[now][dat][tot]!=-)return f[now][dat][tot];
LL ans=,mx=fp?d[now-]:;
for (int i=;i<=mx;i++)
{
if (i==)ans+=dfs(now-,,tot,fp&&i==d[now-]);
else ans+=dfs(now-,,tot+(dat==),fp&&i==d[now-]);
}
if (!fp)f[now][dat][tot]=ans;
return ans;
}
inline LL calc(LL x)
{
if (x<=)return ;
LL xxx=x;
len=;
while (xxx)
{
d[++len]=xxx%;
xxx/=;
}
LL sum=;
for (int i=;i<=d[len];i++)sum+=dfs(len,i,,i==d[len]);
return sum;
}
main()
{
int T=read(),cnt=;
while (T--)
{
r=read();
if (r<l)swap(l,r);
memset(f,-,sizeof(f));
printf("Case %d: %lld\n",++cnt,calc(r));
}
}
LightOj 1032
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