日常不会

链接：https://www.nowcoder.com/acm/contest/85/F

这道题需要想清楚每个图对应的x轴y轴、、

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=2000+10;
const int maax=1e5+10;
int a[maax][5];
char s[maxn][5*maxn],w[maxn][maxn];
int main(){
    int x,y,z,n;
    cin>>x>>y>>z>>n;
    for(int i=0;i<n;i++){
        for(int j=0;j<3;j++)
            cin>>a[i][j];
    }
    for(int i=0;i<y;i++){
        for(int j=0;j<x;j++)
            s[i][j]='.';
    }
    for(int i=0;i<y;i++)
        s[i][x]=' ';
    for(int i=0;i<y;i++){
        for(int j=x+1;j<x+z+1;j++)
            s[i][j]='.';
    }
    for(int i=0;i<n;i++){
        s[y-1-(a[i][1]-1)][a[i][0]-1]='x';
    }
    for(int i=0;i<n;i++){
        s[y-1-(a[i][1]-1)][x+1+(a[i][2]-1)]='x';
    }
    for(int i=0;i<z;i++){
        for(int j=0;j<x;j++)
            w[i][j]='.';
    }
    for(int i=0;i<n;i++){
        w[a[i][2]-1][a[i][0]-1]='x';
    }
    for(int i=0;i<y;i++){
        for(int j=0;j<x+z+1;j++)
            cout<<s[i][j];
        cout<<endl;
    }
    cout<<endl;
    for(int i=0;i<z;i++){
        for(int j=0;j<x;j++)
            cout<<w[i][j];
        cout<<endl;
    }
}

链接：https://www.nowcoder.com/acm/contest/85/H

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;
char s[10][5010];
char ans[5010];
int vis[5010];
int main(){
    int n;
    cin>>n;
    for(int i=0;i<9;i++)
        cin>>s[i];
    for(int i=0;i<9;i++){
        for(int j=0;j<n;j++){
            if(s[i][j]=='o'){
                if(i ==0)ans[j]='F',vis[j]=1;
                if(i==1)ans[j]='E',vis[j]=1;
                if(i==2)ans[j]='D',vis[j]=1;
                if(i==3)ans[j]='C',vis[j]=1;
                if(i==4)ans[j]='B',vis[j]=1;
                if(i==5)ans[j]='A',vis[j]=1;
                if(i==6)ans[j]='G',vis[j]=1;
                if(i==7)ans[j]='F',vis[j]=1;
                if(i==8)ans[j]='E',vis[j]=1;
            }
            if(s[i][j]=='|')ans[j]='|',vis[j]=1;
        }
    }
    for(int i=0;i<n;i++){
        if(vis[i]==1)
            cout<<ans[i];
        else
            cout<<" ";
    }
    cout<<endl;
    return 0;
}

链接：https://www.nowcoder.com/acm/contest/85/

留给以后写吧。。

//I-卡特兰数-组合数学-阶乘逆元快速幂
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn=2*1e5+10;
const int mod=998244353;
ll f[maxn];
void ff(){
    f[0]=1;
    for(int i=1;i<maxn;i++)
        f[i]=(i*f[i-1])%mod;
}
ll poww(ll n,ll m){   //快速幂
    ll ans=1;
    while(m){
        if(m&1)ans=(ans*n)%mod;
        m=m>>1;
        n=(n*n)%mod;
    }
    return ans;
}
ll cc(ll n,ll m){      //组合数
    ll ans=f[n];
    ans=(ans*poww(f[m],mod-2))%mod;
    ans=(ans*poww(f[n-m],mod-2))%mod;
    return ans;
}
int main(){
    int t,n,m,h=0;
    ff();
    cin>>t;
    while(t--){
        cin>>n;
        m=n-1;
        ll ans=(cc(2*n,n)-cc(2*n,n-1)-cc(2*m,m)+cc(2*m,m-1)+2*mod)%mod;
        cout<<"Case #"<<++h<<": "<<ans<<endl;
    }
}

链接：https://www.nowcoder.com/acm/contest/85/G

以后再搞吧。。

　　　　　　　　　　　　　　　　　　　　C. Sonya and Robots

Since Sonya is interested in robotics too, she decided to construct robots that will read and recognize numbers.

Sonya has drawn $\mathrm{nn numbers\; in\; a\; row, aiai is\; located\; in\; the ii-th\; position.\; She\; also\; has\; put\; a\; robot\; at\; each\; end\; of\; the\; row\; (to\; the\; left\; of\; the\; first\; number\; and\; to\; the\; right\; of\; the\; last\; number).\; Sonya\; will\; give\; a\; number\; to\; each\; robot\; (they\; can\; be\; either\; same\; or\; different)\; and\; run\; them.\; When\; a\; robot\; is\; running,\; it\; is\; moving\; toward\; to\; another\; robot,\; reading\; numbers\; in\; the\; row.\; When\; a\; robot\; is\; reading\; a\; number\; that\; is\; equal\; to\; the\; number\; that\; was\; given\; to\; that\; robot,\; it\; will\; turn\; off\; and\; stay\; in\; the\; same\; position.}$

Sonya does not want robots to break, so she will give such numbers that robots will stop before they meet. That is, the girl wants them to stop at different positions so that the first robot is to the left of the second one.

For example, if the numbers $[1,5,4,1,3][1,5,4,1,3] are\; written,\; and\; Sonya\; gives\; the\; number 11 to\; the\; first\; robot\; and\; the\; number 44 to\; the\; second\; one,\; the\; first\; robot\; will\; stop\; in\; the 11-st\; position\; while\; the\; second\; one\; in\; the 33-rd\; position.\; In\; that\; case,\; robots\; will\; not\; meet\; each\; other.\; As\; a\; result,\; robots\; will\; not\; be\; broken.\; But\; if\; Sonya\; gives\; the\; number 44 to\; the\; first\; robot\; and\; the\; number 55 to\; the\; second\; one,\; they\; will\; meet\; since\; the\; first\; robot\; will\; stop\; in\; the 33-rd\; position\; while\; the\; second\; one\; is\; in\; the 22-nd\; position.$

Sonya understands that it does not make sense to give a number that is not written in the row because a robot will not find this number and will meet the other robot.

Sonya is now interested in finding the number of different pairs that she can give to robots so that they will not meet. In other words, she wants to know the number of pairs ($\mathrm{pp, qq),\; where\; she\; will\; give pp to\; the\; first\; robot\; and qq to\; the\; second\; one.\; Pairs\; (pipi, qiqi)\; and\; (pjpj, qjqj)\; are\; different\; if pi\ne pjpi\ne pj or qi\ne qjqi\ne qj.}$

Unfortunately, Sonya is busy fixing robots that broke after a failed launch. That is why she is asking you to find the number of pairs that she can give to robots so that they will not meet.

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; number\; of\; numbers\; in\; a\; row.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1051\le ai\le 105) —\; the\; numbers\; in\; a\; row.}$

Output

Print one number — the number of possible pairs that Sonya can give to robots so that they will not meet.

Examples

input

5
1 5 4 1 3

output

9

input

7
1 2 1 1 1 3 2

output

7

#include<iostream>
#include<algorithm>
#include<vector>
#define ll long long
using namespace std;
vector<int>q;
ll n,w;ll x[100002],y[100002];
int main()
{
        cin>>n;
        ll num=0;ll ans=0; 
        for(int i=0;i<n;i++)
        {
            cin>>w;q.push_back(w);
            x[w]++;
            if(x[w]==1) num++;//计算没有重复出现的数字 
        }
        for(int i=0;i<n;i++)//从左往右的过程 ，后面出现重复的数字会在前面就被遍历到然后就不会重复计算； 
        {
            x[q[i]]--;
            if(x[q[i]]==0) num--;
            if(y[q[i]]==0)
            {
                ans+=num;
                y[q[i]]=1;
            }
        }
        cout<<ans<<endl;
    return 0; 
}

　　　　　　　　　　　　　　　　　　　　　　　　　　　　C. Summarize to the Power of Two

 

　　A sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an is\; called\; good\; if,\; for\; each\; element aiai,\; there\; exists\; an\; element ajaj (i\ne ji\ne j)\; such\; that ai+ajai+aj is\; a\; power\; of\; two\; (that\; is, 2d2d for\; some\; non-negative\; integer dd).}}^{}$

　　For example, the following sequences are good:

• $[5,3,11][5,3,11] (for\; example,\; for a1=5a1=5 we\; can\; choose a2=3a2=3.\; Note\; that\; their\; sum\; is\; a\; power\; of\; two.\; Similarly,\; such\; an\; element\; can\; be\; found\; for a2a2 and a3a3),$
• $[1,1,1,1023][1,1,1,1023],$
• $[7,39,89,25,89][7,39,89,25,89],$
• $[][].$

Note that, by definition, an empty sequence (with a length of $\mathrm{00)\; is\; good.}$

For example, the following sequences are not good:

• $[16][16] (for a1=16a1=16,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two),$
• $[4,16][4,16] (for a1=4a1=4,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two),$
• $[1,3,2,8,8,8][1,3,2,8,8,8] (for a3=2a3=2,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two).$

You are given a sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an.\; What\; is\; the\; minimum\; number\; of\; elements\; you\; need\; to\; remove\; to\; make\; it\; good?\; You\; can\; delete\; an\; arbitrary\; set\; of\; elements.}}^{}$

Input

The first line contains the integer $\mathrm{nn (1\le n\le 1200001\le n\le 120000)\; —\; the\; length\; of\; the\; given\; sequence.}$

The second line contains the sequence of integers ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109).}}^{}$

Output

Print the minimum number of elements needed to be removed from the given sequence in order to make it good. It is possible that you need to delete all $\mathrm{nn elements,\; make\; it\; empty,\; and\; thus\; get\; a\; good\; sequence.}$

Examples

input

6
4 7 1 5 4 9

output

1

input

5
1 2 3 4 5

output

2

input

1
16

output

1
数据太大，用暴力加法肯定超时。需要用一个map来匹配减去其中一个值，在输入中的数字能否找到就行了。不需要管元素是什么，只需要管个数。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 120000 + 10;
int a[maxn];
int p[maxn];
vector<int>V;
map<int,int>M;
int cnt = 0;
void init()
{
    for(int i = 0 ; i<=30 ; i++) p[i] = 1<<i; 
}
int main()
{
    int n;
    scanf("%d",&n);
    init();
    M.clear();
    for(int i = 0 ; i < n ; i++)
    {
        scanf("%d",&a[i]);
        if(!M.count(a[i]))
        {
            M[a[i]] = 1;
        }
        else 
        {
            M[a[i]] ++;
        }
    }
    int ans = 0;
    for(int i = 0 ; i < n ; i++)
    {
        int j;
        for(j = 0 ; j <= 30 ; j++)
        {
            int t = p[j] - a[i];
            if(t == a[i])
            {
                if(M[t]>=2)
                {
                    break;
                }
            }
            else 
            {
                if(M[t])
                {
                    break;
                }
            }
        }
        if(j == 31)
        {
            ans++;
        } 

    }
    cout<<ans<<endl;    
}