简单C语言课设题 （可带写 ）

#define _CRT_SECURE_NO_WARNINGS //VS2019 
#include 




int main() {

	float  Original_value;  //原所得
	float  taxes;              //纳税值
	float  actual_value;   //实际所得

	printf("请输入原始所得：rn");
	scanf("%f", &Original_value);

	if (Original_value < 10000) {
		taxes = Original_value * 0.05f;
		actual_value = Original_value - taxes;

	}
	else if (10000 < Original_value && Original_value <= 20000) {
		taxes = Original_value * 0.1f;
		actual_value = Original_value - taxes;

	}
	else if (20000 < Original_value && Original_value <= 40000) {
		taxes = Original_value * 0.2f;
		actual_value = Original_value - taxes;

	}
	else if (40000 < Original_value && Original_value <= 80000) {
		taxes = Original_value * 0.3f;
		actual_value = Original_value - taxes;

	}
	else {
		taxes = Original_value * 0.35f;
		actual_value = Original_value - taxes;


	}
	printf("应缴纳税额:taxes= %f , 实际所得: actual_value=%f", taxes, actual_value);
}

#include

int main()
{
	char c;
	int letter, blankSpace, number, other;
	letter = blankSpace = number = other = 0;
	while ((c = getchar()) != 'n')
	{
		if (c >= 'a' && c <= 'z' || c >= 'A' && c <= 'Z')
			letter++;
		else if (c >= '0' && c <= '9')
			number++;
		else if (c == ' ')
			blankSpace++;
		else
			other++;
	}

	printf("字母数:%dn空格数:%dn数字数:%dn其他字符:%d", letter, blankSpace, number, other);
	return 0;
}

#define _CRT_SECURE_NO_WARNINGS //VS2019 
#include 
double  fun(double  x) {
	double y=0;
	if (x<1) {
		y = x * 2.0;
	}
	else if (x>=1&&x<=10) {
		y = 2.0 * x - 11.0;
	}
	else {
		y = 2.0 * x + 15.0;
	}
	return y;
}

int main() {
	double i;
	double y = 0;
	scanf("％d",&i);
	y=fun(i);
	printf("函数值为：y=%d",y);

	return 0;
}

//求4*4矩阵最小值
#define _CRT_SECURE_NO_WARNINGS //VS2019
#include "stdio.h"



int main() { 
	int i, j, x, imin;
	int data[4][4] = { 0 };
	printf("请输入矩阵原始数据:(空格分隔)");
	for (i = 0; i < 4; i++) {
		for (j = 0; j < 4; j++) {
			data[i][j] = scanf("%dr", &x);
		}
	}
	printf("输入完毕rn");
	
	for (int i = 0; i < 3; i++)
		for (int j = 0; j < 3; j++)
		{
			int imin = data[0][0];
			if (data[j][i] <= imin)
				imin = data[j][i];
		}
	printf("最小值为：%dn", &imin);
	return 0;

}