For循环
#include<stdio.h>
int main(void)
{
char ch;
for (ch = 'a'; ch <= 'z'; ch++) //一个括号初值终值和增量全给表示了，比while简洁
printf("The ASCII value for %c is %#x. \n", ch, ch); //%#x表示以0x的前缀显示为16进制数
return 0;
}The ASCII value for a is 0x61.
The ASCII value for b is 0x62.
The ASCII value for c is 0x63.
The ASCII value for d is 0x64.
The ASCII value for e is 0x65.
The ASCII value for f is 0x66.
The ASCII value for g is 0x67.
The ASCII value for h is 0x68.
The ASCII value for i is 0x69.
The ASCII value for j is 0x6a.
The ASCII value for k is 0x6b.
The ASCII value for l is 0x6c.
The ASCII value for m is 0x6d.
The ASCII value for n is 0x6e.
The ASCII value for o is 0x6f.
The ASCII value for p is 0x70.
The ASCII value for q is 0x71.
The ASCII value for r is 0x72.
The ASCII value for s is 0x73.
The ASCII value for t is 0x74.
The ASCII value for u is 0x75.
The ASCII value for v is 0x76.
The ASCII value for w is 0x77.
The ASCII value for x is 0x78.
The ASCII value for y is 0x79.
The ASCII value for z is 0x7a.数组
#include<stdio.h>
int main(void)
{
char ch[26]; //数组的容量是26
int i;
printf("请输入26个字母。\n");
for(i = 0; i < 26; i++)
scanf("%c", &ch[i]);
printf("现在将打印刚刚输入的26个字母。\n");
for(i = 0; i < 26; i++)
printf("%c", ch[i]);
printf("\n");
return 0;
}请连续输入26个字母。
qazwsxedcrfvtgbyhnujmikolp
现在将打印刚刚输入的26个字母。
qazwsxedcrfvtgbyhnujmikolp嵌套循环
内外循环无关
#include<stdio.h>
#define R 8
/*
这里可以凸显设定常量的便利，
只要这里变动一下就可改变整个嵌套循环,
而不用一个个进块里头改
*/
int main(void)
{
int row, i;
for(row = 1; row <= R; row++)
{
for(i = 1; i <= row; i++)
{
printf("%c", '$');
}
printf("\n");
}
return 0;
}$
$$
$$$
$$$$
$$$$$
$$$$$$
$$$$$$$
$$$$$$$$外部循环控制内部循环
#include<stdio.h>
#define R 6
int main(void)
{
int row;
char i;
for(row = 1; row <= R; row++)
{
for(i = 'F'; i >= ('G'- row); i--)
{
printf("%c", i);
}
printf("\n");
}
return 0;
}F
FE
FED
FEDC
FEDCB
FEDCBA追加条件改变内部循环初始值
#include<stdio.h>
#define R 6
int main(void)
{
int row, sum;
char i;
sum = 0;
for(row = 0; row < R; row++)
{
sum += row;
for(i = 'A' + sum; i <= ('A' + sum + row); i++)
{
printf("%c", i);
}
printf("\n");
}
return 0;
}A
BC
DEF
GHIJ
KLMNO
PQRSTU多个内部循环
基础版
#include<stdio.h>
#define R 10
int main(void)
{
int row, sum;
char i, j, k;
for(row = 0; row < R; row++)
{
for(k = R - row - 1; k > 0; k--)
{
printf(" ");
}
for(i = 'A'; i <= ('A' + row); i++)
{
printf("%c", i);
}
for(j = 'A' + row - 1; j >= 'A'; j--)
{
printf("%c", j);
}
printf("\n");
}
return 0;
} A
ABA
ABCBA
ABCDCBA
ABCDEDCBA
ABCDEFEDCBA
ABCDEFGFEDCBA
ABCDEFGHGFEDCBA
ABCDEFGHIHGFEDCBA
ABCDEFGHIJIHGFEDCBA添加交互与判断版
使用if和goto，直到用户输入符合要求的内容为止不断跳出提示
#include<stdio.h>
int main(void)
{
int row, sum, r;
char i, j, k, h;
test:
printf("Please enter a CAPITAL letter: ");
scanf("%s", &h);
if (sizeof(h) > 1)
{
goto test;
}
else if(h < 65)
{
goto test;
}
else if(h > 90)
{
goto test;
}
else
{
goto ct;
}
ct:
printf("(If you enter a string with a capital initial,\nthe program will process only this initial.)\n\n");
r = h - 64;
for(row = 0; row < r; row++)
{
for(k = r - row - 1; k > 0; k--)
{
printf(" ");
}
for(i = 'A'; i <= ('A' + row); i++)
{
printf("%c", i);
}
for(j = 'A' + row - 1; j >= 'A'; j--)
{
printf("%c", j);
}
printf("\n");
}
return 0;
}Please enter a CAPITAL letter: a
Please enter a CAPITAL letter: c
Please enter a CAPITAL letter: 7
Please enter a CAPITAL letter: +-+
Please enter a CAPITAL letter: &*
Please enter a CAPITAL letter: jij
Please enter a CAPITAL letter: 74kj
Please enter a CAPITAL letter: k7
Please enter a CAPITAL letter: Fast
(If you enter a string with a capital initial,
the program will process only this initial.)
A
ABA
ABCBA
ABCDCBA
ABCDEDCBA
ABCDEFEDCBA制表
#include<stdio.h>
int main(void)
{
int r, i;
double a[100];
printf("将打印一个表格，显示所输入数的1到3次方。\n");
printf("请输入表格的行数：");
scanf("%d", &r);
printf("请输入%d个数：", r);
for(i = 0; i < r; i++)
{
scanf("%lf", &a[i]);
}
printf("1次方 2次方 3次方\n");
for(i = 0; i < r; i++)
{
printf("%-15.2f%-15.2f%-15.2f\n", a[i], a[i] * a[i], a[i] * a[i] * a[i]);
printf("\n");
}
return 0;
}将打印一个表格，显示所输入数的1到3次方。
请输入表格的行数：3
请输入3个数：74 85 96
1次方 2次方 3次方
74.00 5476.00 405224.00
85.00 7225.00 614125.00
96.00 9216.00 884736.00
倒序输出
数字
#include<stdio.h>
int main(void)
{
int a[8], i;
printf("请输入8个整数：");
for(i = 0; i < 8; i++)
{
scanf("%d", &a[i]);
}
printf("现将倒序打印这8个整数：");
for(i = 7; i >= 0; i--)
{
printf("%d", a[i]);
}
return 0;
}请输入8个整数：1 2 3 4 5 6 7 8
现将倒序打印这8个整数：87654321字母
#include<stdio.h>
#include<string.h>
int main(void)
{
char name[20];
int len, i;
printf("Please enter your name (up to 20 characters):\n");
scanf("%s", name);
len = strlen(name);
printf("Now something will happen:\n");
for(i = len - 1; i >= 0; i--)
{
printf("%c", name[i]);
}
return 0;
}Please enter your name(up to 20 characters):
SMITH
Now something will happen:
HTIMS级数
序列1
$$ \sum_{n=1}^{\infty} \frac{1}{n} \to\ \infty $$
序列2
$$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}=\log (2) \approx 0.69315 $$
#include<stdio.h>
#include<math.h>
int main(void)
{
int m, i, j;
double t1 = 0.0, t2 = 0.0;
printf("请输入项数：");
scanf("%d", &m);
while (m > 0)
{
t1 = 0.0;
for(i = 1; i <= m; i++)
{
t1 = t1 + 1.0 / (double)i;
}
t2 = 0.0;
for(j = 1; j <= m; j++)
{
t2 = t2 + pow(-1.0, (double)j + 1.0) / (double)j;
}
printf("序列1之和（%d项为止）是：%f。\n", m, t1);
printf("序列2之和（%d项为止）是：%f。\n", m, t2);
printf("请输入项数：");
scanf("%d", &m);
}
printf("你输入了非正数，程序结束。\n");
return 0;
}请输入项数：100
序列1之和（100项为止）是：5.187378。
序列2之和（100项为止）是：0.688172。
请输入项数：1000
序列1之和（1000项为止）是：7.485471。
序列2之和（1000项为止）是：0.692647。
请输入项数：10000
序列1之和（10000项为止）是：9.787606。
序列2之和（10000项为止）是：0.693097。
请输入项数：100000
序列1之和（100000项为止）是：12.090146。
序列2之和（100000项为止）是：0.693142。
请输入项数：1000000
序列1之和（1000000项为止）是：14.392727。
序列2之和（1000000项为止）是：0.693147。
请输入项数：10000000
序列1之和（10000000项为止）是：16.695311。
序列2之和（10000000项为止）是：0.693147。
请输入项数：-100
你输入了非正数，程序结束。Do while循环
注意while外部指定初始值，内部指定迭代的结构
#include<stdio.h>
#include<math.h>
int main(void)
{
int a[8], i;
printf("计算2的前8次方。\n");
for(i = 0; i < 8; i++)
{
a[i] = pow(2, i + 1);
}
i = 0;
do{
printf("%d\n", a[i]);
i++;
}while(i < 8);
return 0;
}计算2的前8次方。
2
4
8
16
32
64
128
256渐化式
#include<stdio.h>
int main(void)
{
double a[8], b[8];
int i;
printf("请输入8个数。\n");
for(i = 0; i < 8; i++)
{
scanf("%lf", &a[i]);
}
for(i = 0; i < 8; i++)
{
printf("%-9.2f", a[i]);
}
printf("\n");
b[0] = a[0];
for(i = 0; i < 8; i++)
{
b[i] = b[i - 1] + a[i]; //S(n) = S(n - 1) + a(n)
printf("%-9.2f", b[i]);
}
return 0;
}请输入8个数。
1 2 3 4 5 6 7 8
1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00
1.00 3.00 6.00 10.00 15.00 21.00 28.00 36.00单利与复利
#include<stdio.h>
#include<math.h>
int main(void)
{
double a, b;
int i = 1;
do{
a = 100 + 10.0 * i;
b = 100 * pow(1.05, i);
i++;
}while(a > b);
printf("经过%d年，对本金100元，5%%复利(b)的本息合计会大于10%%单利(a)。\n", --i);
printf("此时a的本息合计是%.2f元，b的本息合计是%.2f元。\n", a, b);
return 0;
}经过27年，对本金100元，5%复利(b)的本息合计会大于10%单利(a)。
此时a的本息合计是370.00元，b的本息合计是373.35元。牛吃草问题
（1）草地100坪，生草速度8坪/单位时间，吃草速度10坪/单位时间，求吃完需要的时间
#include<stdio.h>
int main(void)
{
double a, b;
int i = 1;
do{
a = 100 + 8 * i;
b = 10 * i;
i++;
}while(a > b);
printf("t = %d\n", --i);
return 0;
}t = 50（2）草地5坪，第n周吃掉n坪的草，剩下的草翻倍成长，求草场面积超过150的耗时
#include<stdio.h>
int main(void)
{
double a = 5.0;
int i = 1;
do{
a = (a - i) * 2;
printf("In the #%d week, the area will be %.1fpy.\n", i, a);
i++;
}while(a <= 150.0);
printf("In the #%d week, the area will exceed 150.0py.\n", --i);
return 0;
}In the #1 week, the area will be 8.0py.
In the #2 week, the area will be 12.0py.
In the #3 week, the area will be 18.0py.
In the #4 week, the area will be 28.0py.
In the #5 week, the area will be 46.0py.
In the #6 week, the area will be 80.0py.
In the #7 week, the area will be 146.0py.
In the #8 week, the area will be 276.0py.
In the #8 week, the area will exceed 150.0py.