最大公約數(shù)與最小公倍數(shù)
//輾轉(zhuǎn)相除法求公約公倍
#include <stdio.h>
int gcd(int a, int b) {
return (a % b == 0) ? b : gcd(b, a % b);
}
int main() {
int m, n;
scanf("%d %d", &m, &n);
int ans = gcd(m, n);
printf("%d %d\n", ans, m * n / ans);
return 0;
}
?字符串分類統(tǒng)計(jì)
#include <stdio.h>
#include <string.h>
int
main(void)
{
char str[200];//聲明一個(gè)數(shù)組用于存儲(chǔ)輸入的字符
int letter = 0, number = 0, space = 0, other, n;
gets(str); //獲取輸入并存儲(chǔ)到數(shù)組里,gets會(huì)把輸入的換行符\n丟棄
n = strlen (str); //獲取輸入的字符長(zhǎng)度
for(int i = 0; i < n; i++)//for循環(huán)可以聲明變量同時(shí)初始化,多個(gè)聲明用 , 逗號(hào)隔開(kāi)
{
if ((str[i] >= 'A' && str[i] <= 'Z') || (str[i] >= 'a' && str[i] <= 'z'))
letter += 1;
if (str[i] == ' ')
space += 1;
if (str[i] >= '0' && str[i] <= '9')
number += 1;
}
other = n - space - number - letter; //把整個(gè)字符長(zhǎng)度減去字母數(shù)、空格、數(shù)字得出其他字符的數(shù)量
printf("%d %d %d %d", letter, number, space, other);
return 0;
}
?Sn的公式求和
#include<stdio.h>
int main(){
int n, i, sum = 0;
scanf("%d", &n);
for(i = 0; i < n; i++){
sum *= 10;
sum += 2 * (i+1);
}
printf("%d", sum);
return 0;
}
階乘求和
#include <stdio.h>
int main(void)
{
int n;
long long Sn = 0, j;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
j = 1;
for(int m = 0; m < i; m++)
j *= (m + 1);
Sn = Sn + j;
}
printf("%lld", Sn); //long long的格式指定符為%lld
return 0;
}
求和訓(xùn)練?
?
#include<stdio.h>
int main(void)
{
int An = 0, Bn = 0;
float Cn = 0;
int a, b, c;
float i; //這里i的聲明類型也可以為int,但如果這樣,下面有些地方就要做出改變
scanf("%d %d %d", &a, &b, &c);
for (i = 1; a >= i; i++)
{
An = i + An;
}
for (i = 1; b >= i; i++)
{
Bn = i * i + Bn;
}
for ( i = 1; c >= i; i++)
{
Cn = 1 / i + Cn; /*如果上面i的定義不為浮點(diǎn)型而是整型,這里的1就應(yīng)該變?yōu)?.0,否則1/i的結(jié)果就不是浮點(diǎn)型,小數(shù)點(diǎn)后面位數(shù)的數(shù)值會(huì)直接被忽略,這里也是導(dǎo)致程序結(jié)果雖然保留兩位小數(shù),但小數(shù)點(diǎn)后的位數(shù)卻是零的緣故。*/
}
printf("%.2f", An + Bn + Cn);
return 0;
}
?水仙花數(shù)判斷
#include<stdio.h>
int main(){
int i;
for(i=100;i<=999;i++){
if(i==(i/100)*(i/100)*(i/100)+((i%100)/10)*((i%100)/10)*((i%100)/10)+(i%10)*(i%10)*(i%10))
printf("%d\n",i);
}
return 0;
}
完數(shù)的判斷
#include<stdio.h>
#define max 1001
int main() {
int n,i,j,a[max],l;
scanf("%d",&n);
for(i=1; i<=n; i++) {
int tem=0,k=0; //每次開(kāi)始新一輪的內(nèi)循環(huán)之前重置
for(j=1; j<i; j++) {
if(i%j==0) {//滿足條件
tem+=j;//儲(chǔ)存因子之和
a[k++]=j;//儲(chǔ)存因子
}
}
if(tem==i) {//滿足條件 格式輸出
printf("%d its factors are",i);
for(l=0;l<k;l++){
printf(" %d",a[l]);
}
printf("\n");
}
}
return 0;
}
有規(guī)律的數(shù)列求和
#include<stdio.h>
int main(){
double Sn=0,an=0,n,i,t,a=2.0,b=1.0;//a為分子,b為分母,an為單項(xiàng)和值,Sn為總和
scanf("%lf",&n);
Sn+=a/b;//第一項(xiàng)不在規(guī)律中,在循環(huán)前先加入
for(i=1;i<n;i++){
t=a;
a=a+b;//將前一項(xiàng)的分子加分母給后一項(xiàng)的分子
b=t;//將前一項(xiàng)的分子給后一項(xiàng)的分母
an=a/b;
Sn+=an;
}
printf("%0.2lf",Sn);
return 0;
}
自由下落的距離計(jì)算
#include<stdio.h>
int main()
{
double m, h, ans;
int n;
scanf("%lf%d",&m,&n);
h = m / (1 << n);
ans = m + (m - h * 2) * 2;
printf("%.2lf %.2lf", h, ans);
return 0;
}
猴子吃桃的問(wèn)題?
文章來(lái)源:http://www.zghlxwxcb.cn/news/detail-794408.html
# include <stdio.h>
int main()
{
int sum=1,N;
scanf("%d",&N);
while(--N)
sum=(sum+1)*2;
printf("%d",sum);
return 0;
}
迭代法求平方根?
?文章來(lái)源地址http://www.zghlxwxcb.cn/news/detail-794408.html
#include <stdio.h>
#include <math.h>
int main()
{
float left,right,mid;
float fleft, fright,fmid;
left=-10.;
right=10.;
mid=(left+right)/2;
fmid=2*mid*mid*mid-4*mid*mid+3*mid-6;
while(fabs(fmid)>1e-6)
{
fleft=2*left*left*left-4*left*left+3*left-6;
fright=2*right*right*right-4*right*right+3*right-6;
if(fleft*fmid>0)
left=mid;
else
right=mid;
mid=(left+right)/2;
fmid=2*mid*mid*mid-4*mid*mid+3*mid-6;
}
printf("%.2f\n",mid);
return 0;
}
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