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  • codeforces493B

    Vasya and Wrestling

     CodeForces - 493B 

    Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.

    When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.

    If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.

    Input

    The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).

    The following n lines contain integer numbers ai (|ai| ≤ 109ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.

    The techniques are given in chronological order.

    Output

    If the first wrestler wins, print string "first", otherwise print "second"

    Examples

    Input
    5
    1
    2
    -3
    -4
    3
    Output
    second
    Input
    3
    -1
    -2
    3
    Output
    first
    Input
    2
    4
    -4
    Output
    second

    Note

    Sequence x  =  x1x2... x|x| is lexicographically larger than sequence y  =  y1y2... y|y|, if either |x|  >  |y| and x1  =  y1,  x2  =  y2, ... ,  x|y|  =  y|y|, or there is such number r(r  <  |x|, r  <  |y|), that x1  =  y1,  x2  =  y2,  ... ,  xr  =  yr and xr  +  1  >  yr  +  1.

    We use notation |a| to denote length of sequence a.

    sol:直接按照题意O(n)模拟就可以了,注意和会爆int

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    inline ll read()
    {
        ll s=0;
        bool f=0;
        char ch=' ';
        while(!isdigit(ch))
        {
            f|=(ch=='-'); ch=getchar();
        }
        while(isdigit(ch))
        {
            s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
        }
        return (f)?(-s):(s);
    }
    #define R(x) x=read()
    inline void write(ll x)
    {
        if(x<0)
        {
            putchar('-'); x=-x;
        }
        if(x<10)
        {
            putchar(x+'0'); return;
        }
        write(x/10);
        putchar((x%10)+'0');
        return;
    }
    #define W(x) write(x),putchar(' ')
    #define Wl(x) write(x),putchar('
    ')
    const int N=200005;
    int n,a[N],b[N];
    int main()
    {
        int i,opt=0;
        ll A=0,B=0,Last;
        R(n);
        for(i=1;i<n;i++)
        {
            ll x=read();
            if(x>0)
            {
                a[++*a]=x; A+=1ll*x;
            }
            else
            {
                x=-x; b[++*b]=x; B+=1ll*x;
            }
        }
        R(Last);
        if(Last>0)
        {
            a[++*a]=Last; A+=Last;
        }
        else
        {
            b[++*b]=-1*Last; B+=-1*Last;
        }
        if(A!=B)
        {
            (A>B)?puts("first"):puts("second");
        }
        else
        {
            for(i=1;i<=max(*a,*b);i++) if(a[i]!=b[i])
            {
                (a[i]>b[i])?puts("first"):puts("second");
                return 0;
            }
            if(Last>0) puts("first");
            else puts("second");
        }
        return 0;
    }
    /*
    input
    3
    1000000000
    1000000000
    1000000000
    output
    first
    */
    View Code
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  • 原文地址:https://www.cnblogs.com/gaojunonly1/p/10631585.html
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