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;
}