圈

题目地址：http://www.51cpc.com/web/problem.php?id=4286

解法 1：

 1 #include<iostream>
 2 #include<map>
 3 using namespace std;
 4 
 5 #define LL long long
 6 map<LL, LL> imap;
 7 int n;
 8 
 9 int main()
10 {
11     ios::sync_with_stdio(false);
12     while(cin>>n) {
13         LL a,b;
14         for(int i=0; i<n; i++) {
15             cin>>a>>b;
16             imap[a] += b;
17         }
18         
19         LL ans=0, last=0;
20         map<LL, LL>::iterator iter = imap.begin();
21         for(; iter!=imap.end(); iter++) {
22             map<LL, LL>::iterator p = ++iter;
23             iter--;
24             
25             LL one=iter->first, two=max(iter->second, last);
26             if(p == imap.end()) {
27                 ans = one+1;
28                 two = two/4 + (two%4==0?0:1);
29                 while(two > 1) {
30                     two = two/4 + (two%4==0?0:1);
31                     ans++;
32                 }
33                 break;
34             }
35             
36             last=two;
37             while(one < p->first) {
38                 one++;
39                 last = last/4 + (last%4==0?0:1);
40                 if(last == 1) break;
41             }
42         }
43         cout<<ans<<endl;
44         imap.clear();
45     }
46     
47     return 0;
48 }

解法 2：机智的解法

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cmath>
 4 #include <algorithm>
 5 using namespace std;
 6 const int MAXN=2e5+5;
 7 struct Node
 8 {
 9     int w,cnt;
10 }data[MAXN];
11 int main()
12 {
13     int n;
14     while(scanf("%d",&n)!=EOF)
15     {
16         int ans=0;
17         for(int i=0;i<n;i++)
18         {
19             cin>>data[i].w>>data[i].cnt;
20             int tot=1;
21             for(int j=data[i].w+1;;j++)
22             {
23                 tot*=4;
24                 if(tot>=data[i].cnt)
25                 {
26                     ans=max(ans,j);break;
27                 }
28             }
29         }
30         cout<<ans<<endl;
31     }
32     ret

解法 3：这个简直超神了。。看到这个瞬间就清醒了，就是求4的多少次，特判下0的情况就行；

 1 #include<cstdio>
 2 #include<cmath>
 3 int N,R=0,T,K,A;
 4 main(){
 5     scanf("%d",&N);
 6     while(N--){
 7         scanf("%d%d",&K,&A);
 8         A=ceil(log2(A)/2);
 9         K+=1>A?1:A;
10         R=R<K?K:R;
11     }
12     printf("%d",R);
13 }