真是状况百出的一次CF啊……
最终还Unrated了，你让半夜打cf 的我们如何释怀（中途茫茫多的人都退场了）……虽说打得也不好……
在这里写一下这一场codeforces的解题报告，A-E的 题目及AC代码，部分题目有简单评析，代码还算清晰，主要阅读代码应该不难以理解。
A. Factory
One industrial factory is reforming working plan. The director suggested to set a mythical detail production norm. If at the beginning of the day there were x details in the factory storage, then by the end of the day the factory has to produce 
The board of directors are worried that the production by the given plan may eventually stop (that means that there will be а moment when the current number of details on the factory is divisible by m).
Given the number of details a on the first day and number m check if the production stops at some moment.
The first line contains two integers a and m (1 ≤ a, m ≤ 105).
Print "Yes" (without quotes) if the production will eventually stop, otherwise print "No".
1 5
No
3 6
Yes
那就用大小为m的一个vis数组来记录当前余数是否被用过，然后模拟，每次记录当前余数，如果余数变成0了输出Yes，如果余数到达了曾经有过的余数位置，那么就会以此为一个循环永远循环下去，那么我们break，输出No
Code：
#include <cmath>
#include <cctype>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
int main()
{
int a,m; cin>>a>>m;
int vis[100086]={0};
while(1)
{
if(a==0){cout<<"Yes";return 0;}
if(vis[a]==1){cout<<"No";return 0;}
vis[a]=1;
a=(a+a)%m;
}
return 0;
}
这不知道算不算凸包，反正记录最大最小的x和y，然后相减获得最小矩形长宽，取两者较长边平方即可。
Code：
#include <cmath>
#include <cctype>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int inf=(int)1e9+10086;
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
bool cmp(const int a, const int b)
{
return a > b;
}
int main()
{
int n; cin>>n;
int up=-inf,down=inf,left=inf,right=-inf;
for(int i=0;i<n;i++)
{
int x,y; cin>>x>>y;
if(x<left) left=x;
if(x>right) right=x;
if(y<down) down=y;
if(y>up) up=y;
}
int len=max(up-down,right-left);
cout<<(long long)len*len;
return 0;
}
C. Bits
Let's denote as 
You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l ≤ x ≤ r, and
The first line contains integer n — the number of queries (1 ≤ n ≤ 10000).
Each of the following n lines contain two integers li, ri — the arguments for the corresponding query (0 ≤ li ≤ ri ≤ 1018).
For each query print the answer in a separate line.
3 1 2 2 4 1 10
1 3 7
The binary representations of numbers from 1 to 10 are listed below:
110 = 12
210 = 102
310 = 112
410 = 1002
510 = 1012
610 = 1102
710 = 1112
810 = 10002
910 = 10012
1010 = 10102
这题是给一个范围（L是左边界，R是有边界）问你在这个范围内哪个数载二进制下1的数量是最多的（有多个解请输出最小数）。
也就是，要二进制的1尽量多，还要求尽量小，那就从低位开始把0变成1呗
那么我们就从左边界开始，从低位向高位按位或（0变成1，1也还是1）1，直到比R大停止，输出前一个（即比R小的最后一个数）。
Code：
#include <cmath>
#include <cctype>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
int main()
{
int cases=0;
scanf("%d",&cases);
for(int _case=1;_case<=cases;_case++)
{
ll l,r,t,p=1; cin>>l>>r;
for(ll i=0;i<63;i++)
{
ll t=l|p;
if(t>r)break;
l=t,p<<=1;
//cout<<t<<endl;
}
cout<<l<<endl;
}
return 0;
}
D. Maximum Value
You are given a sequence a consisting of n integers. Find the maximum possible value of 
The first line contains integer n — the length of the sequence (1 ≤ n ≤ 2·105).
The second line contains n space-separated integers ai (1 ≤ ai ≤ 106).
Print the answer to the problem.
3 3 4 5
2
int n=0; cin>>n;
for(int i=0;i<n;i++)
scanf("%d",&num[i]);
sort(num,num+n,cmp);
int modmax=0;
for(int i=0; num[i]>modmax; i++)
for(int j=i+1;num[j]>modmax && j<n; j++)
update( num[i] % num[j]);
cout<<modmax;
return 0;这样的东西……TLE的飞起……
想了想就不能暴力啊，暴力的话剪枝也没用
Code：
#include <cmath>
#include <cctype>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,a[200048]={0},ans=0;
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define update(x) ans=(ans<(x)?x:ans);
int main()
{
scanf("%d",&n);
for(int i=0;i<n;++i) scanf("%d",&a[i]);
sort(a,a+n);
for(int i=0;i<n-1;++i)
if(i==0||a[i]!=a[i-1])
{
int j=a[i]+a[i],p;
while(j <= a[n-1])
{
p = lower_bound(a,a+n,j)-a;
if(p > 0) update(a[p-1] % a[i]);
j+=a[i];
}
update(a[n-1] % a[i]);
}
printf("%d\n",ans);
return 0;
}
DIV2全场只有一个人（joker99）出了E，看了下代码暂时囫囵吞了下，贴一下代码等日后学习下。
Code：
#include <cmath>
#include <cctype>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
const int size = 2 * 1000 * 1000 + 10;
const int ssize = 21;
char buf[size];
char nbuf[size];
int n=0, m=0;
int pwr[ssize][size];
void combine(int* tg, int* a, int* b, int shift)
{
for (int i = 0; i < n; i++)
{
if (a[i] < shift) tg[i] = a[i];
else tg[i] = b[a[i] - shift] + shift;
}
}
int main()
{
scanf("%s", buf); n = strlen(buf);
scanf("%d", &m);
for (int i = 0; i < m; i++)
{
int k, d;
scanf("%d%d", &k, &d);
int num = n - k + 1, cur = 0;
for (int j = 0; j < d; j++)
{
int p = j;
while (p < k)
{
pwr[0][p] = cur++;
p += d;
}
}
for (int j = k; j < n; j++) pwr[0][j] = j;
int lim = 0, vl = 1;
while (vl <= num)
{
vl *= 2;
lim++;
}
for (int j = 1; j < lim; j++)
combine(pwr[j], pwr[j - 1], pwr[j - 1], (1 << (j - 1)));
for (int j = 0; j < n; j++)
{
int ps = j;
int vl = num;
int sh = 0;
for (int h = lim - 1; h >= 0; h--)
if (vl >= (1 << h))
{
if (ps >= sh) ps = pwr[h][ps - sh] + sh;
vl -= (1 << h);
sh += (1 << h);
}
nbuf[ps] = buf[j];
}
for (int j = 0; j < n; j++) buf[j] = nbuf[j];
printf("%s\n", buf);
}
return 0;
}