牛客多校训练AFJ(签到)

题目链接https://ac.nowcoder.com/acm/contest/881/A（单调栈）

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+;
typedef long long ll;
using namespace std;
int L1[*maxn],L2[*maxn];
int a[*maxn],b[*maxn];
int main()
{
   int n;
   while(~scanf("%d",&n))
   {
       for(int t=;t<=n;t++)
       {
           scanf("%d",&a[t]);
    }
    for(int t=;t<=n;t++)
    {
        scanf("%d",&b[t]);
    }
    stack<int>s1,s2;
    for(int t=;t<=n;t++)
    {
        if(s1.empty())
        {
            L1[t]=;
            s1.push(t);
            continue;
        }
        else
        {
            while(!s1.empty()&&a[s1.top()]>a[t])
            {
                s1.pop();
            }
           if(s1.empty())
           {
            L1[t]=;
            s1.push(t);
            continue;
           }
           else
           {
               L1[t]=s1.top();
               s1.push(t);
               continue;
           }
        }
    }
    for(int t=;t<=n;t++)
    {
        if(s2.empty())
        {
            L2[t]=;
            s2.push(t);
            continue;
        }
        else
        {
            while(!s2.empty()&&b[s2.top()]>b[t])
            {
                s2.pop();
            }
           if(s2.empty())
           {
            L2[t]=;
            s2.push(t);
            continue;
           }
           else
           {
               L2[t]=s2.top();
               s2.push(t);
               continue;
           }
        }
    }
    int k=n+;
    for(int t=;t<=n;t++)
    {
        //cout<<L1[k]<<" "<<L2[k]<<endl;
        if(L1[t]!=L2[t]){
            k=t;
            break;
        }
    }
    printf("%d\n",k-);
 
   }
   return ;
}

题目链接：https://ac.nowcoder.com/acm/contest/881/F（思维）

三角形面积的22倍

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+;
typedef long long ll;
using namespace std;
int main()
{  
 
   long double x0,y0,x1,y1,x2,y2;
   long double a,b,c,p,S;
   long double ans;
   while(cin>>x0>>y0>>x1>>y1>>x2>>y2)
   {
        a=sqrtl((x0-x1)*(x0-x1)+(y0-y1)*(y0-y1));
        b=sqrtl((x0-x2)*(x0-x2)+(y0-y2)*(y0-y2));
        c=sqrtl((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
        //cout<<a<<" "<<b<<" "<<c<<endl;
        p=(a+b+c)/;
        S=sqrtl(p*(p-a)*(p-b)*(p-c));
        S=S*;
        ans=S;
        printf("%.0Lf\n",ans);
   }
 
   return ;
}

海伦公式版没过 不知道原因

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+;
typedef long long ll;
using namespace std;
int main()
{ 
 
   long double x0,y0,x1,y1,x2,y2;
   long double a,b,c,p,S;
   long double ans;
   while(cin>>x0>>y0>>x1>>y1>>x2>>y2)
   {
     a=sqrtl((x0-x1)*(x0-x1)+(y0-y1)*(y0-y1));
     b=sqrtl((x0-x2)*(x0-x2)+(y0-y2)*(y0-y2));
     c=sqrtl((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
     //cout<<a<<" "<<b<<" "<<c<<endl;
     p=(a+b+c)/;
     S=sqrtl(p*(p-a)*(p-b)*(p-c));
     S=S*;
     ans=S;
     printf("%.0Lf\n",ans);
   }
 
   return ;
}

题目链接：https://ac.nowcoder.com/acm/contest/881/J（大数）

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
using namespace std; 
 
const int maxn = ; 
 
struct bign{
    int d[maxn], len; 
 
    void clean() { while(len >  && !d[len-]) len--; } 
 
    bign()          { memset(d, , sizeof(d)); len = ; }
    bign(int num)   { *this = num; }
    bign(char* num) { *this = num; }
    bign operator = (const char* num){
        memset(d, , sizeof(d)); len = strlen(num);
        for(int i = ; i < len; i++) d[i] = num[len--i] - '';
        clean();
        return *this;
    }
    bign operator = (int num){
        char s[]; sprintf(s, "%d", num);
        *this = s;
        return *this;
    } 
 
    bign operator + (const bign& b){
        bign c = *this; int i;
        for (i = ; i < b.len; i++){
            c.d[i] += b.d[i];
            if (c.d[i] > ) c.d[i]%=, c.d[i+]++;
        }
        while (c.d[i] > ) c.d[i++]%=, c.d[i]++;
        c.len = max(len, b.len);
        if (c.d[i] && c.len <= i) c.len = i+;
        return c;
    }
    bign operator - (const bign& b){
        bign c = *this; int i;
        for (i = ; i < b.len; i++){
            c.d[i] -= b.d[i];
            if (c.d[i] < ) c.d[i]+=, c.d[i+]--;
        }
        while (c.d[i] < ) c.d[i++]+=, c.d[i]--;
        c.clean();
        return c;
    }
    bign operator * (const bign& b)const{
        int i, j; bign c; c.len = len + b.len;
        for(j = ; j < b.len; j++) for(i = ; i < len; i++)
            c.d[i+j] += d[i] * b.d[j];
        for(i = ; i < c.len-; i++)
            c.d[i+] += c.d[i]/, c.d[i] %= ;
        c.clean();
        return c;
    }
    bign operator / (const bign& b){
        int i, j;
        bign c = *this, a = ;
        for (i = len - ; i >= ; i--)
        {
            a = a* + d[i];
            for (j = ; j < ; j++) if (a < b*(j+)) break;
            c.d[i] = j;
            a = a - b*j;
        }
        c.clean();
        return c;
    }
    bign operator % (const bign& b){
        int i, j;
        bign a = ;
        for (i = len - ; i >= ; i--)
        {
            a = a* + d[i];
            for (j = ; j < ; j++) if (a < b*(j+)) break;
            a = a - b*j;
        }
        return a;
    }
    bign operator += (const bign& b){
        *this = *this + b;
        return *this;
    } 
 
    bool operator <(const bign& b) const{
        if(len != b.len) return len < b.len;
        for(int i = len-; i >= ; i--)
            if(d[i] != b.d[i]) return d[i] < b.d[i];
        return false;
    }
    bool operator >(const bign& b) const{return b < *this;}
    bool operator<=(const bign& b) const{return !(b < *this);}
    bool operator>=(const bign& b) const{return !(*this < b);}
    bool operator!=(const bign& b) const{return b < *this || *this < b;}
    bool operator==(const bign& b) const{return !(b < *this) && !(b > *this);} 
 
    string str() const{
        char s[maxn]={};
        for(int i = ; i < len; i++) s[len--i] = d[i]+'';
        return s;
    }
}; 
 
istream& operator >> (istream& in, bign& x)
{
    string s;
    in >> s;
    x = s.c_str();
    return in;
} 
 
ostream& operator << (ostream& out, const bign& x)
{
    out << x.str();
    return out;
}
int main()
{
    bign x,y,a,b;
    while(cin>>x>>a>>y>>b)
    {
        if(x*b==y*a)
        {
            puts("=");
        }
        else if(x*b<y*a)
        {
            puts("<");
        }
        else
        {
            puts(">");
        }
    }
}