1+22+333+……的C语言代码

以下是输出1 22 333 ……的C语言代码：

#include <stdio.h>

int main() {
    int n = 10; // 输出前10个数
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            printf("%d", i); // 输出i
        }
        printf(" "); // 输出空格
    }
    return 0;
}

输出结果为：

1 22 333 4444 55555 666666 7777777 88888888 999999999 10101010101010101010

相关问题

使用函数实现求下列值++1！+1！+2！+1！+2！+3！+……+1！+2！+3！+……+n！用C语言

以下是使用函数实现求下列值1！+2！+3！+4！+…+10！的C语言代码：

#include <stdio.h>
#include <stdlib.h>

int Factor(int n) //计算n的阶乘
{
    int result = 1;
    for (int i = 1; i <= n; i += 1) {
        result *= i;
    }
    return result;
}

int FactorSum(int n) //计算1！+2！+3！+...+n！
{
    int sum = 0;
    for (int i = 1; i <= n; i += 1) {
        sum += Factor(i);
    }
    return sum;
}

int main()
{
    int n = 10;
    int result = FactorSum(n);
    printf("1!+2!+3!+...+%d! = %d\n", n, result);
    system("pause");
    return 0;
}

使用函数实现求下列值 1！+1！+2！+1！+2！+3！+……+1！+2！+3！+……+n！用C语言

以下是使用函数实现求下列值1！+1！+2！+1！+2！+3！+……+1！+2！+3！+……+n！的C语言代码：

#include <stdio.h>
#include <stdlib.h>

int Factor(int n) //计算n的阶乘
{
    int result = 1;
    for (int i = 1; i <= n; i += 1) {
        result *= i;
    }
    return result;
}

int FactorSum(int n) //计算1!+2!+3!+...+n!
{
    int sum = 0;
    for (int i = 1; i <= n; i += 1) {
        for (int j = 1; j <= i; j += 1) {
            sum += Factor(j);
        }
    }
    return sum;
}

int main() {
    int n = 5; //n为5
    int result = FactorSum(n);
    printf("result = %d\n", result); //输出结果
    system("pause");
    return 0;
}