c语言代码如下：include<stdio.h> int main(){double sum=0,x=1;while(sum<5){sum+=(1/x);x++;} printf("出满足不等式1+1/2+1/3+...+1/n≥5的最小n值为%.0lf\n",x-1);return 0;}
下面是使用循环来求n个数的最小值的C程序示例: #include <stdio.h> int main() { int n; printf("请输入要比较的数字个数:"); scanf("%d", &n); int num, min; printf("请输入第1个数:"); scanf("%d", &min); for (int i = 1; i < n; i++) { printf("请输入第%d个数:", i+...
} printf("这%d个数中的最小个位数字是:%d\n", n, min_digit); return 0; } ``` 这段代码首先获取用户输入的数字n,然后通过取余运算(`n % 10`)获取这个数字的个位数。然后,如果这个新的个位数比之前的个位数更小,就更新最小值。这个过程会一直重复,直到n小于10为止。最后,打印出最小的个位数。©...
for(int i=1;;i+=2)if((i%3==2)&&(i%5==3)&&(i%7==4)){ printf("%d\n",i);break;}
C语言输入n和n个数字找出里面的最小值 #include <stdio.h>#include<stdlib.h>intmain() {intarr[101];intn,i;intflag =1001; scanf("%d",&n);for(i=0;i<n;i++) { scanf("%d",&arr[i]);if(arr[i]<flag) flag=arr[i]; } printf("%d",flag);return0;...
include <stdio.h> int main() { int n, m;int s = 0;scanf("%d", &m);for (n = 1; ; n++) { s += 2 * n - 1;if (s >= m)break;} printf("%d\n", s == m ? n + 1 : n);return 0;}
define N 20 include<stdio.h> void main(){ int a[N]; //定义一个数组来存储输入的n个数，N的值可修改 int n，i;int min; //用来存储最小的那个值 scanf("%d",&n);printf("请输入%d个数",n);//接收输入的n个数 for(i=0;i<n;i++){ scanf("%d",&a[i]);} //求...
include <stdio.h>int main(void){ int min=9999, max=0; //设最小值不小于0,最大值不大于9999 int i; printf("%s\n", "Please input a series of integer numbers,-1 to end input :"); do { scanf( "%d", &i); if( i < min) min=i; if( i > m...
include <stdio.h>int main(int argc, char const *argv[]){int a , n, min;scanf("%d",&n);scanf("%d",&min);for (int i = 1; i < n; ++i){scanf("%d",&a);if (a < min){min = a;}}printf("min=%d\n", min);return 0;} ...
printf("你要输入几个数：");scanf("%d",&n);printf("输入%d个数:",n);int s[n];scanf("%d",&s[0]);int min=s[0],max=s[0];for(int i=1;i<n;i++){scanf("%d",&s[i]);if(min>s[i])min=s[i];if(max