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  • UVa1585 Score 题解

    UVa1585 题目原地址

    本体思路重点在于分数的统计及重置。
    由题面可得输入的是一个小于80个字符的字符串,故使用一个字符数组存储。

    更新 2020/3/1 11:52
    傻了,uva评测太慢代码挂上就先去摸鱼,结果一看wa了。看了下代码,只写了主功能函数。给定的输入有多个kase...
    代码更新如下(ifdef内的是调试代码,可去掉):

    #include <stdio.h>
    #include <cstdlib>
    
    int getScore() {
    	int quantity = 0;
    	int score = 0;
    	char inputs[80] = {};
    
    	scanf("%s", inputs);
    	#ifdef DEBUG
    	printf("Current string: %s
    ", inputs);
    	#endif
    
    	for(int i = 0; i <= 80; i++){
    		if(inputs[i] == 'O') {
    			quantity++;
    			score += quantity;
    			#ifdef DEBUG
    			printf("Current quantity of O: %d
    ", quantity);
    			printf("Score: %d
    
    ", score);
    			#endif
    		}
    		if(inputs[i] == 'X') {
    			quantity = 0;
    			#ifdef DEBUG
    			printf("Current character is X.
    ");
    			printf("Score: %d
    
    ", score);
    			#endif
    		}
    	}
    	
    	return score;
    }
    
    int main(void){
    
    	#ifdef DEBUG
    	freopen("data.in", "r", stdin);
    	#endif
    
    	int *scores;
    	
    	int kase_quantity = 0;
    	scanf("%d", &kase_quantity);
    	// create a array to save scores of each kase
    	scores = (int*)calloc(kase_quantity, sizeof(int));
    
    	for(int i = 0; i <= kase_quantity; i++) {
    		int score = getScore();
    		scores[i] = score;
    	}
    	for(int i = 0; i < kase_quantity; i++) {
    		printf("%d
    ", scores[i]);
    	}
    
    	// free memory used by scores array
    	free(scores);
    
    	return 0;
    }
    

    废话不多说,代码如下:

    #include <stdio.h>
    
    int main(void){
    	char inputs[80];
    	int quantity = 1;
    	int score;
    	
    	scanf("%s", inputs);
    	for(int i = 0; i <= 80; i++){
    		if(inputs[i] == 'O') {
    			score += quantity;
    			quantity++;
    		}
    		if(inputs[i] == 'X') {
    			quantity = 1;
    		}
    	}
    
    	printf("%d
    ", score);
    	
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/kozumi/p/12609314.html
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