Codeforces Round #450 (Div. 2) ABCD

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
3
1 1
-1 -1
2 -1
output
Yes
input
4
1 1
2 2
-1 1
-2 2
output
No
input
3
1 2
2 1
4 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 }
View Code

 

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 }
View Code

 

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
1
1
output
1
input
5
5 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 }
View Code

 

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 }
View Code

 

posted @ 2017-12-12 11:16  hk_lin  阅读(623)  评论(0)  编辑  收藏  举报