1032 - Fast Bit Calculations 数位dp

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本文介绍了一种计算从0到N所有整数中相邻1出现次数的方法。通过将整数转换为二进制并使用动态规划算法进行计算，提供了一个有效的解决方案。

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1032 - Fast Bit Calculations

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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 1adjacent 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

Output for Sample Input

2147483647

Case 1: 0

Case 2: 2

Case 3: 12

Case 4: 13

Case 5: 13

Case 6: 14

Case 7: 16106127360

PROBLEM SETTER: MOHIUL ALAM PRINCE

SPECIAL THANKS: JANE ALAM JAN (MODIFIED DESCRIPTION, DATASET)

把数转化成2进制然后按位dp就好

ACcode：

#include <cstdio>
#include <cstring>
#define ll long long
ll dp[37][37][2];///0--->都没有 1--->前面有1
int data[36];
int cnt=1;
ll dfs(int len,int num,int flag,int limit){
    if(!len) return num;
    if(!limit&&dp[len][num][flag]!=-1)return dp[len][num][flag];
    int ed=limit?data[len]:1;
    ll ans=0;
    for(int i=0;i<=ed;++i)
        if(flag)ans+=dfs(len-1,num+(i==1),i==1,limit&&i==ed);
        else ans+=dfs(len-1,num,i==1,limit&&i==ed);
    return limit?ans:dp[len][num][flag]=ans;
}
void doit(){
    ll n;
    scanf("%lld",&n);
    int len=0;
    while(n){
        data[++len]=n%2;
        n>>=1LL;
    }
    printf("Case %d: %lld\n",cnt++,dfs(len,0,0,1));
}
int main(){
    int loop;memset(dp,-1,sizeof(dp));
    scanf("%d",&loop);
    while(loop--)doit();
    return 0;
}