100000569 - 《算法笔记》2.5小节——C/C++快速入门->数组

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本文精选了多个C/C++编程中数组操作的经典案例，包括有序插入、数组逆置、杨辉三角生成、字符串解密与比较、逆序输出数组、Fibonacci数列求解、冒泡排序、矩阵转置及字符串最大值查找等，深入浅出地讲解了数组的基本用法和高效编程技巧。

《算法笔记》2.5小节——C/C++快速入门->数组
26039 Problem A 习题6-4 有序插入

#include <stdio.h>

int main() {
	int a[10];
	for (int i = 0; i < 10; i++) {
		scanf("%d", &a[i]);
	}
	int temp = a[9];
	int j;
	for (j = 8; temp<a[j]; j--) {
		a[j + 1] = a[j];
	}
	a[j+1] = temp;

	for (int i = 0; i < 10; i++) {
		printf("%d\n",a[i]);
	}
	return 0;
}

26040 Problem B 习题6-5 数组元素逆置

#include <stdio.h>

int main() {
	int a[10];
	for (int i = 0; i < 10; i++) {
		scanf("%d", &a[i]);
	}

	for (int i = 0; i < 5;i++) {
		int temp = a[i];
		a[i] = a[10 - 1 - i];
		a[10 - 1 - i] = temp;
	}

	for (int i = 0; i < 10; i++) {
		printf("%d\n",a[i]);
	}
	return 0;
}

26041 Problem C 习题6-6 杨辉三角

#include <stdio.h>

int main() {
	const int n = 10;
	int count;
	scanf("%d", &count);
	int a[n][2*n] = { 0 };
	for (int i = 0; i < count; i++) {
		a[i][0] = 1;
	}
	for (int i = 1; i < count; i++){
		for (int j = 1; j <= i; j++) {
			a[i][j] = a[i - 1][j-1 ] + a[i - 1][j];
		}
	}
	for (int i = 0; i < count; i++) {
		for (int j = 0; j <= i; j++) {
			if (i != j) printf("%d ", a[i][j]);
			else if (i == j && i < count - 1) printf("%d\n", a[i][j]);
			else printf("%d", a[i][j]);
		}
	}

	return 0;
}

26044 Problem D 习题6-12 解密

#include <stdio.h>
#include <string.h>

int main() {
	char a[20];
	scanf("%s", a);
	for (int i = 0; i < strlen(a); i++) {
		if (a[i] >= 65 && a[i] <= 90) {
			a[i] = 26 - (a[i] - 64) + 65;
		}else if (a[i] >= 97 && a[i] <= 122) {
			a[i] = 26 - (a[i] - 96) + 97;
		}
	}
	printf("%s", a);
}

26045 Problem E 习题6-13 字符串比较

#include <stdio.h>
#include <string.h>

int main() {
	char a[20], b[20];
	gets(a);
	gets(b);

	int len = strlen(a) < strlen(b) ? strlen(a) : strlen(b);
	bool equal = true;

	for (int i = 0; i < len; i++) {
		if (a[i] != b[i]) {
			printf("%d", a[i] - b[i]);
			equal = false;
			break;
		}
	}
	
	if (equal == true) printf("%d", 0);

	return 0;
}

26049 Problem F 例题6-1 逆序输出数组元素

#include <stdio.h>
#include <string.h>

int main() {
	int a[10];
	for (int i = 0; i < 10; i++) {
		scanf("%d", &a[i]);
	}
	for (int i = 0; i < 5; i++) {
		int temp = a[i];
		a[i] = a[9 - i];
		a[9 - i] = temp;
	}
	for (int i = 0; i < 10; i++) {
		printf("%d", a[i]);
		if (i != 9) printf("\n");
	}
	return 0;

}

26050 Problem G 例题6-2 数组求解Fibonacci数列问题

#include <stdio.h>
#include <string.h>

int main() {
	int a[20] = { 1,1 };
	
	for (int i = 2; i < 20; i++) {
		a[i] = a[i - 1] + a[i - 2];
	}
	for (int i = 0; i < 20; i++) {
		printf("%d\n", a[i]);
	}
	
	return 0;
}

26051 Problem H 例题6-3 冒泡排序

#include <stdio.h>
#include <string.h>

int main() {
	int a[10];
	
	for (int i = 0; i < 10; i++) {
		scanf("%d", &a[i]);
	}
	for (int i = 0; i < 10; i++) {
		for (int j = 0; j < 10 - i - 1; j++) {
			if (a[j] > a[j + 1]) {
				int temp = a[j];
				a[j] = a[j + 1];
				a[j + 1] = temp;
			}
		}
	}
	for (int i = 0; i < 10; i++) {
		printf("%d\n", a[i]);
	}
	return 0;
}

26052 Problem I 例题6-4 矩阵转置

#include <stdio.h>
#include <string.h>

int main() {
	int a[2][3], b[3][2];
	for (int i = 0; i < 2; i++) {
		for (int j = 0; j < 3; j++) {
			scanf("%d", &a[i][j]);
		}
	}
	for (int i = 0; i < 2; i++) {
		for (int j = 0; j < 3; j++) {
			b[j][i] = a[i][j];
		}
	}
	for (int i = 0; i < 3; i++) {
		for (int j = 0; j < 2; j++) {
			printf("%d ", b[i][j]);
		}
		printf("\n");
	}
	return 0;
}

26055 Problem J 例题6-9 字符串求最大值

#include <stdio.h>
#include <string.h>

int main() {
	char a[3][10];
	for (int i = 0; i < 3; i++) {
		gets(a[i]);
	}
	int max = 0;
	for (int i = 1; i < 3; i++) {
		if (strcmp(a[max], a[i]) < 0) max = i;
	}
	printf("%s", a[max]);

	return 0;
}