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  • CodeForces 493B Vasya and Wrestling 【模拟】

    B. Vasya and Wrestling
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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"

    Sample test(s)
    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.

    一定要注意是不是会越界啊!

    !。!

    代码:

    #include <stdio.h>
    #include <string.h>
    #define M 250050
    typedef unsigned long long LL;
    
    int a[M], b[M];
    
    int main(){
    	int n;
    	while(~scanf("%d", &n)){
    		int temp, i, j;
    		int flag = -1, la, lb;
    		la = lb = 0;
    		LL sum1, sum2;
    		sum1 = sum2 = 0;
    		for(i = 0; i < n; i ++){
    			scanf("%d", &temp);
    			if(temp > 0){
    				sum1 +=temp;
    				flag = 1;
    				a[la++] = temp;
    			}
    			else {
    				temp = -temp;
    				sum2 += temp;
    				flag = 2;
    				b[lb++] = temp;
    			}
    		}
    		if(sum1 > sum2){
    			printf("first
    ");
    		}
    		else if(sum1 < sum2){
    			printf("second
    ");
    		}
    		else{
    			i = 0; j = 0;
    			int ok = 0;
    			while(i<la&&j<lb){
    				if(a[i] == b[j]){
    					++i; ++j;
    				}
    				else{
    					if(a[i] > b[j]) puts("first");
    					else puts("second");
    					return 0;
    				}
    			}
    			if(la >lb){
                    printf("first
    "); return 0;
    			}
    			if(la < lb){
                    printf("second
    ");
                    return 0;
    			}
    			if(flag == 1) printf("first
    ");
    			else printf("second
    ");
            }
    	}
    	return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/slgkaifa/p/7384621.html
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