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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; }
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5746509.html