这次还是能看的0 0，没出现一题掉分情况。
QAQ前两次掉分还被hack了0 0,两行清泪。
A. Find Extra One
You have n distinct points on a plane, none of them lie on OY axis. Check that there is a point after removal of which the remaining points are located on one side of the OY axis.
Input
The first line contains a single positive integer n (2?≤?n?≤?105).
The following n lines contain coordinates of the points. The i-th of these lines contains two single integers xi and yi (|xi|,?|yi|?≤?109,xi?≠?0). No two points coincide.
Output
Print "Yes" if there is such a point, "No" — otherwise.
You can print every letter in any case (upper or lower).
Examples
input
31 1-1 -12 -1
output
Yes
input
41 12 2-1 1-2 2
output
No
input
31 22 14 60
output
Yes
题意：给你n个平面上的点，能否去掉一个使得所有点在y轴一侧。给出的点不会在y轴上。
题解：水题，统计下x>0 和 x<0的数字个数就好了。
1 #include<bits/stdc++.h>
2 #define clr(x) memset(x,0,sizeof(x))
3 #define clr_1(x) memset(x,-1,sizeof(x))
4 #define mod 1000000007
5 #define LL long long
6 #define INF 0x3f3f3f3f
7 using namespace std;
8 int main()
9 {
10 int n,left,right,x,y;
11 left=right=0;
12 scanf("%d",&n);
13 for(int i=1;i<=n;i++)
14 {
15 scanf("%d%d",&x,&y);
16 if(x>0)
17 right++;
18 else
19 left++;
20 }
21 if(right<=1 || left<=1)
22 printf("Yes\n");
23 else
24 printf("No\n");
25 return 0;
26 }
B. Position in Fraction
You have a fraction 
. You need to find the first occurrence of digit c into decimal notation of the fraction after decimal point.
Input
The first contains three single positive integers a, b, c (1?≤?a?<?b?≤?105, 0?≤?c?≤?9).
Output
Print position of the first occurrence of digit c into the fraction. Positions are numbered from 1 after decimal point. It there is no such position, print -1.
Examples
input
1 2 0
output
2
input
2 3 7
output
-1
题意：给出一个分数$ \frac{a}{b} $ ，看c在该分数小数点后几位。
题解：$ \frac{a}{b} $化为最简$ \frac{a‘}{b‘} $后，我们模拟除法列竖式求答案的过程。由于列竖式的过程中余下的数总比b‘小，所以最多b‘个不同余数，列竖式的过程中一旦出现相同的余数则是进入了循环。所以我们最多做b‘次除法就能求出c是否存在于该分数小数点后以及c的位置。
1 #include<bits/stdc++.h>
2 #define clr(x) memset(x,0,sizeof(x))
3 #define clr_1(x) memset(x,-1,sizeof(x))
4 #define mod 1000000007
5 #define LL long long
6 #define INF 0x3f3f3f3f
7 using namespace std;
8 int gcd(int a,int b)
9 {
10 int c;
11 while(b)
12 {
13 c=a%b;
14 a=b;
15 b=c;
16 }
17 return a;
18 }
19 int main()
20 {
21 int a,b,c,d,e,i;
22 scanf("%d%d%d",&a,&b,&c);
23 d=gcd(a,b);
24 a/=d;
25 b/=d;
26 d=a/b;
27 a-=d*b;
28 a*=10;
29 for(i=1;i<=100000;i++)
30 {
31 d=a/b;
32 a-=d*b;
33 if(d==c)
34 break;
35 a*=10;
36 }
37 if(i>100000)
38 printf("-1\n");
39 else
40 printf("%d\n",i);
41 return 0;
42 }
C. Remove Extra One
You are given a permutation p of length n. Remove one element from permutation to make the number of records the maximum possible.
We remind that in a sequence of numbers a1,?a2,?...,?ak the element ai is a record if for every integer j (1?≤?j?<?i) the following holds:aj?<?ai.
Input
The first line contains the only integer n (1?≤?n?≤?105) — the length of the permutation.
The second line contains n integers p1,?p2,?...,?pn (1?≤?pi?≤?n) — the permutation. All the integers are distinct.
Output
Print the only integer — the element that should be removed to make the number of records the maximum possible. If there are multiple such elements, print the smallest one.
Examples
input
11
output
1
input
55 1 2 3 4
output
5
Note
In the first example the only element can be removed.
题意:给出1~n的一个排列，要求你找出一个数，在排列中去掉它以后符合条件的数最多，如果有多个则选最小的那个。若ai符合条件，对于任意1≤j<i，aj<ai。
题解：那么我们写个bit来记录数列前面小于ai的个数，set来求取大于ai的第一个数。如果ai前面有i-2个数小于ai，那么用set的lonwer_bound找到大于ai的第一个数p，并且该数p对于答案的贡献num[p]++。注意如果ai前面有i-1个数小于ai，那么ai对于答案的贡献num[ai]--。然后从小到大找贡献最大的那个。
1 #include<bits/stdc++.h>
2 #define clr(x) memset(x,0,sizeof(x))
3 #define clr_1(x) memset(x,-1,sizeof(x))
4 #define mod 1000000007
5 #define LL long long
6 #define INF 0x3f3f3f3f
7 using namespace std;
8 const int N=1e5+10;
9 int bits[N];
10 int num[N];
11 int n,m,T,k,p;
12 set<int> pt;
13 set<int>::iterator it;
14 void add(int i,int x)
15 {
16 while(i<=n)
17 {
18 bits[i]+=x;
19 i+=(i &(-i));
20 }
21 return ;
22 }
23 int sum(int i)
24 {
25 int res=0;
26 while(i>0)
27 {
28 res+=bits[i];
29 i-=(i &(-i));
30 }
31 return res;
32 }
33 int main()
34 {
35 scanf("%d",&n);
36 for(int i=1;i<=n;i++)
37 {
38 scanf("%d",&p);
39 if((p-1>0 && sum(p-1)==i-2) || (p-1==0 && i==2))
40 {
41 it=pt.lower_bound(p);
42 num[*it]++;
43 }
44 if((p-1>0 && sum(p-1)==i-1) || (p-1==0 && i==1))
45 num[p]--;
46 pt.insert(p);
47 add(p,1);
48 }
49 int maxn=num[1];
50 k=1;
51 for(int i=2;i<=n;i++)
52 {
53 if(maxn<num[i])
54 {
55 maxn=num[i];
56 k=i;
57 }
58 }
59 printf("%d\n",k);
60 return 0;
61 }
D. Unusual Sequences
Count the number of distinct sequences a1,?a2,?...,?an (1?≤?ai) consisting of positive integers such that gcd(a1,?a2,?...,?an)?=?x and 
. As this number could be large, print the answer modulo 109?+?7.
gcd here means the greatest common divisor.
Input
The only line contains two positive integers x and y (1?≤?x,?y?≤?109).
Output
Print the number of such sequences modulo 109?+?7.
Examples
input
3 9
output
3
input
5 8
output
0
题意：让你找出符合条件的序列数，并mod 1e9+7。序列需满足gcd（a1~n）=x,sum(a1~n）=y。
题解：首先把一个数p分解成几个数成组成的序列有$ 2^{p-1} $，用组合的隔板法轻易可证。x能整除y说明存在这样的序列，反之不存在。然后d=y/x，将d分解成质数乘积，算算不同质数的个数以及具体是哪些质数，容斥一下就能求出答案了。
1 #include<bits/stdc++.h>
2 #define clr(x) memset(x,0,sizeof(x))
3 #define clr_1(x) memset(x,-1,sizeof(x))
4 #define mod 1000000007
5 #define LL long long
6 #define INF 0x3f3f3f3f
7 using namespace std;
8 const int N=1e5+10;
9 int prime[N],inf[N],pcnt;
10 void primer(int n)
11 {
12 pcnt=0;
13 for(int i=2;i<=n;i++)
14 {
15 if(!inf[i])
16 {
17 prime[++pcnt]=i;
18 }
19 for(int j=1;j<=pcnt;j++)
20 {
21 if(prime[j]>n/i) break;
22 inf[prime[j]*i]=1;
23 if(i%prime[j]==0) break;
24 }
25 }
26 return ;
27 }
28 LL quick_pow(LL x,LL n)
29 {
30 LL res=1;
31 while(n)
32 {
33 if(n&1)
34 res=(res*x)%mod;
35 n>>=1;
36 x=(x*x)%mod;
37 }
38 return res;
39 }
40 LL x,y,d,n,k;
41 int all;
42 LL num[20];
43 int cnt;
44 LL ans,p;
45 int main()
46 {
47 primer(100000);
48 scanf("%lld%lld",&x,&y);
49 if(y%x!=0)
50 {
51 printf("0\n");
52 return 0;
53 }
54 y/=x;
55 d=y;
56 cnt=0;
57 for(int i=1;(LL)prime[i]*prime[i]<=d;i++)
58 {
59 n=(LL)prime[i];
60 if(d%n==0)
61 {
62 num[++cnt]=n;
63 while(d%n==0) d/=n;
64 }
65 }
66 if(d!=1) num[++cnt]=d;
67 ans=0;
68 all=(1<<cnt)-1;
69 for(int i=0,j,k,ok;i<=all;i++)
70 {
71 j=i;
72 k=0;
73 p=1;
74 ok=1;
75 while(j)
76 {
77 k++;
78 if(j&1) p*=num[k],ok=-ok;
79 j>>=1;
80 }
81 ans=(ans+quick_pow(2,y/p-1)*ok)%mod;
82 }
83 printf("%lld\n",(ans%mod+mod)%mod);
84 return 0;
85 }