HDU 5491.The Next【2015 ACM/ICPC Asia Regional Hefei Online】【方法】【9月30】

最新推荐文章于 2019-04-07 16:38:20 发布

Mr丶Gao

最新推荐文章于 2019-04-07 16:38:20 发布

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分类专栏：【思维练习】文章标签： ACM HDU C++

本文链接：https://blog.youkuaiyun.com/a995549572/article/details/48832997

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本文介绍了一种高效算法，用于解决寻找大于给定整数D的下一个WYH数的问题，即二进制中1的个数位于指定范围内的最小整数。文章提供了两种实现思路，一种为直接搜索法，另一种则是优化后的算法，后者避免了超时问题。

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The Next

Problem Description

Let

L denote the number of 1s in integer

D’s binary representation. Given two integers

S1 and

S2, we call

D a WYH number if

S1≤L≤S2.
With a given

D, we would like to find the next WYH number

Y, which is JUST larger than

D. In other words,

Y is the smallest WYH number among the numbers larger than

D. Please write a program to solve this problem.

Input

The first line of input contains a number

T indicating the number of test cases (

T≤300000).
Each test case consists of three integers

S1, and

S2, as described above. It is guaranteed that

0≤D<231 and

D is a WYH number.

Output

For each test case, output a single line consisting of “Case #X: Y”.

X is the test case number starting from 1.

Y is the next WYH number.

Sample Input

Sample Output

Case #1: 12
Case #2: 25
Case #3: 17

这个题直接爆搜找是会超时的。方法：找到二进制1的个数比s1小的数，然后最小位往前是0的变1，直到1的个数等于s1。代码如下：

#include<iostream>
#include<cstdio>
using namespace std;
int main(){
    int T,kase=1;
    scanf("%d",&T);
    while(T--){
        long long D,s1,s2,cnt;
        scanf("%lld%lld%lld",&D,&s1,&s2);
        for(long long i=D+1;;i++){
            long long n=i,f[100]={0},t=0;
            cnt=0;
            while(n){
                f[t++]=n%2;
                n/=2;
                if(f[t-1]==1) cnt++;
            }
            if(cnt>=s1&&cnt<=s2){//符合题意
                printf("Case #%d: %lld\n",kase++,i);
                break;
            }
            else if(cnt<s1){//可以变
                t=0;
                while(cnt<s1){
                    if(f[t]==0){
                        cnt++;
                        f[t]=1;
                    }
                    t++;
                }
                long long sum=0,x=1;
                for(int j=0;j<100;j++){
                    sum+=(f[j]*x);
                    x*=2;
                }
                printf("Case #%d: %lld\n",kase++,sum);
                break;
            }
        }
    }
    return 0;
}

然后是爆搜程序，可以拿去对拍：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int main(){
    int T,kase=1;
    scanf("%d",&T);
    while(T--){
        int D,s1,s2;
        scanf("%d%d%d",&D,&s1,&s2);
        for(int i=D+1;;i++){
            int n=i,cnt=0;
            while(n){
                if(n%2==1) cnt++;
                n/=2;
            }
            if(cnt>=s1&&cnt<=s2){
                printf("Case #%d: %d\n",kase++,i);
                break;
            }
        }
    }
    return 0;
}