题目链接:
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
/************************************************
┆ ┏┓ ┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃ ┃ ┆
┆┃ ━ ┃ ┆
┆┃ ┳┛ ┗┳ ┃ ┆
┆┃ ┃ ┆
┆┃ ┻ ┃ ┆
┆┗━┓ ┏━┛ ┆
┆ ┃ ┃ ┆
┆ ┃ ┗━━━┓ ┆
┆ ┃ AC代马 ┣┓┆
┆ ┃ ┏┛┆
┆ ┗┓┓┏━┳┓┏┛ ┆
┆ ┃┫┫ ┃┫┫ ┆
┆ ┗┻┛ ┗┻┛ ┆
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + ‘0‘);
putchar(‘\n‘);
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=2e5+10;
const int maxn=2e3+14;
const double eps=1e-12;
LL ans,f[40];
void dfs(LL p,LL q,LL time,int num)
{
if(max(p,0LL)==q)
{
ans=min(ans,time);
return ;
}
else if(max(p,0LL)<q)
{
LL temp=q-max(0LL,p);
ans=min(ans,time+max(temp-num,0LL));
return ;
}
int l=1;
while(p-(f[l]-1)>q)l++;
dfs(p-(f[l]-1),q,time+l,num);
if(l>1)dfs(p-(f[l-1]-1),q,time+l,num+1);
}
inline void solve(LL p,LL q)
{
ans=inf;
dfs(p,q,0,0);
printf("%lld\n",ans);
}
inline void Init()
{
f[0]=1;
For(i,1,35)f[i]=f[i-1]*2;
}
int main()
{
int t;
LL p,q;
read(t);
Init();
while(t--)
{
read(p);read(q);
solve(p,q);
}
return 0;
}
原文:http://www.cnblogs.com/zhangchengc919/p/5746509.html