UVa 1225 Digit Counting

最新推荐文章于 2025-09-09 11:36:41 发布

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本文详细解析了一段复杂代码逻辑的实现过程，包括初始化数组、计数及数据处理方法。通过实例展示了如何优化初始化过程以提高效率，并在后续数据处理中灵活运用逻辑判断，实现对特定数据集的有效检索与分析。代码示例涵盖了数组操作、循环结构及条件判断等关键编程技术，旨在提升读者在复杂数据处理场景下的编程能力。

3-3

此题代码较为繁琐，前半部分初始化一整个数组为 "1234567891011....99989999" ，以便多组数据输入时直接查找即可，不必费时初始化；后半部分即为计数部分，代码如下：

#include <stdio.h>

char s[50000];

int main(void)
{
	int T, N, count = 1, i = 0, find;
	int n0,n1,n2,n3,n4,n5,n6,n7,n8,n9;
	n0=n1=n2=n3=n4=n5=n6=n7=n8=n9=0;
	while (count / 10 == 0) {
		s[i++] = count;
		count++;
	}
	while (count / 100 == 0) {
		s[i++] = count / 10;
		s[i++] = count % 10;
		count++;
	}
	while (count / 1000 == 0) {
		s[i++] = count / 100;
		s[i++] = (count/10) % 10;
		s[i++] = count % 10;
		count++;
	}
	while (count / 10000 == 0) {
		s[i++] = count / 1000;
		s[i++] = (count/100) % 10;
		s[i++] = (count/10) % 10;
		s[i++] = count % 10;
		count++;
	}
	scanf("%d", &T);
	getchar();
	while (T--) {
		scanf("%d", &N);
		getchar();
		if (N / 10 == 0)
			find = N - 1;
		else if (N / 100 == 0)
			find = 2*N - 10;
		else if (N / 1000 == 0)
			find = 3*N - 109;
		else
			find = 4*N - 1108;
		for (i = find; i >= 0; i--)
			switch (s[i]) {
				case 0:
					n0++;
					break;
				case 1:
					n1++;
					break;
				case 2:
					n2++;
					break;
				case 3:
					n3++;
					break;
				case 4:
					n4++;
					break;
				case 5:
					n5++;
					break;
				case 6:
					n6++;
					break;
				case 7:
					n7++;
					break;
				case 8:
					n8++;
					break;
				case 9:
					n9++;
					break;
			}
		printf("%d %d %d %d %d %d %d %d %d %d\n",n0,n1,n2,n3,n4,n5,n6,n7,n8,n9);
		n0=n1=n2=n3=n4=n5=n6=n7=n8=n9=0;
	}
	return 0;
}