编写程序实现以下功能
- 编写程序,打印99乘法表
- 将一面额为10元倍数的整钱(
- 输入一行字符,统计其中单词的个数。各单词之间用空格分隔,空格数可以是多个。
输入输出示例
Input words: The sum of the odd numbers.
Count = 6
程序1:九九乘法表(下三角)
#include
int main()
{
for(int i=1; i9; i++)
{
for(int j=1; ji; j++)
{
printf("%d*%d=%.2d\t",j,i,j*i);
}
printf("\n");
}
printf("\n");
return 0;
}
程序2:换零钱
思路:将所有符合要求的每一种零钱零钱数量用数组存放起来
#include
#include
int main(void)
{
int x,y,i = 0, z,min,j,m;
int a[100], f[93],s[47],t[19];
int sum_money;
printf("请输入需要转换的钱[面额为10元倍数的整钱();
scanf("%d",&sum_money);
int b = sum_money-7,c = (sum_money - 6)/2, d = ( sum_money - 3)/5;
printf("所有方案如下:\n");
for(x = 1; x b; x++)
{
for(y = 1; y c; y++)
{
for(z = 1; z d; z++)
{
if(x + 2*y + 5*z == sum_money)
{
printf("¥1:%2d ¥2:%2d ¥5:%2d\n",x,y,z);
a[i] = (x+y+z);
f[i] = x;
s[i] = y;
t[i] = z;
i++;
}
}
}
}
min = a[0];
x = 0, y = 0,z = 0;
for(j = 1; j < i; j++)
{
if(a[j] < min)
{
m = min;
min = a[j];
a[j] = m;
x = j;
y = j;
z = j;
}
}
printf("零钱数量最少为:%d\n方案为:\n1¥:%d 2¥ : %d 5¥:%d\n",min,f[x],s[y],t[z]);
}
程序3:单词统计
#include
void main() {
char c;
int word = 0;
printf("Please input a sentence:");
for( ; (c=getchar())!='\n';)
{
if(c==' ')continue;
else for( ; (c=getchar())!='\n'; )
{
if(c==' ')
{
word++;
break;
}
}
}
printf("This sentence have %d words",word+1);
}
Original: https://www.cnblogs.com/iforeverhz/p/16255988.html
Author: iforeverhz
Title: 实验
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