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  • BNU 51002 BQG's Complexity Analysis

    模拟

    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<vector>
    #include<algorithm>
    using namespace std;
    
    const int maxn=100+10;
    char t1[maxn],t2[maxn];
    int g1,g2,h1,h2;
    
    
    
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            g1=g2=h1=h2=0;
            scanf("%s%s",t1,t2);
    
            if(t1[2]=='n') g1=1;
            if(t1[3]=='^')
            {
                g1=0;
                for(int i=4;;i++)
                {
                    if(t1[i]>='0'&&t1[i]<='9') g1=g1*10+t1[i]-'0';
                    else break;
                }
            }
    
            int pos=-1;
            int len=strlen(t1);
            for(int i=0;t1[i];i++)
            {
                if(i+3<len&&t1[i]=='l'&&t1[i+1]=='o'&&t1[i+2]=='g'&&t1[i+3]=='n')
                {
                    h1=1;
                    pos=i+6;
                    break;
                }
            }
            if(pos!=-1)
            {
                int sum=0;
                for(int i=pos;i<len;i++)
                {
                    if(t1[i]>='0'&&t1[i]<='9') sum=sum*10+t1[i]-'0';
                    else break;
                }
                h1=max(h1,sum);
            }
    
    
            if(t2[2]=='n') g2=1;
            if(t2[3]=='^')
            {
                g2=0;
                for(int i=4;;i++)
                {
                    if(t2[i]>='0'&&t2[i]<='9') g2=g2*10+t2[i]-'0';
                    else break;
                }
            }
    
             pos=-1;
             len=strlen(t2);
            for(int i=0;t2[i];i++)
            {
                if(i+3<len&&t2[i]=='l'&&t2[i+1]=='o'&&t2[i+2]=='g'&&t2[i+3]=='n')
                {
                    h2=1;
                    pos=i+6;
                    break;
                }
            }
            if(pos!=-1)
            {
                int sum=0;
                for(int i=pos;i<len;i++)
                {
                    if(t2[i]>='0'&&t2[i]<='9') sum=sum*10+t2[i]-'0';
                    else break;
                }
                h2=max(h2,sum);
            }
    
          //  printf("%d %d
    ",g1,h1);
          //  printf("%d %d
    ",g2,h2);
    
          if(g1<g2) printf("First
    ");
          else if(g1>g2) printf("Second
    ");
          else
          {
              if(h1<h2) printf("First
    ");
              else if(h1>h2) printf("Second
    ");
              else printf("Both
    ");
          }
    
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zufezzt/p/5077781.html
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