、绘制流程图，实现sum=1/1!+1/2!+1/3！-+……+1/10!

以下是C语言实现求和的流程图：

st=>start: 开始
in=>inputoutput: 输入x和n的值
op1=>operation: 初始化sum、zi、mu为1
op2=>operation: 计算zi=pow(x,i)
op3=>operation: 计算mu=mu*j
op4=>operation: 计算sum=sum+zi/mu
cond=>condition: i<=n?
out=>inputoutput: 输出sum
e=>end: 结束

st->in->op1->op2->op3->op4->cond
cond(yes)->op2
cond(no)->out->e

以下是C语言实现求和的代码：

#include <stdio.h>
#include <math.h>

int main() {
    int i, j;
    int x, n;
    double sum = 1, zi = 1, mu = 1;

    scanf("%d %d", &x, &n);

    for (i = 1; i <= n; i++) {
        zi = pow(x, i);
        for (j = 1; j <= i; j++)
            mu = mu * j;
        sum = sum + zi / mu;        mu = 1;
    }

    printf("%lf", sum);
    printf("\n");

    return 0;
}