无法调用可以单独运行的自定义函数

问题遇到的现象和发生背景

在上实验课，然后teacher要求写题目，题目挺简单

问题相关代码

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

//using namespace std;

void t5()
{
    for (int i = 9; i > 0; i--)
    {
        if (i<=5)
        {
            for (int temp = 5-i; temp > 0; temp--)
            {
                printf_s("  ");
            }
            for (int temp=i*2-1; temp > 0; temp--)
            {
                printf_s("0 ");
            }
        }
        else
        {
            for (int temp = i - 5; temp > 0; temp--)
            {
                printf_s("  ");
            }
            for (int temp=(10-i)*2-1; temp > 0; temp--)
            {
                printf_s("0 ");
            }
        }
        printf_s("\n");
    }
    return 0;
}

void t4()
{
    for (int x = 1; x <= 9; x++)
    {
        for (int y = 1; y <= x; y++)
        {
            int temp = x * y;
            printf_s("%d * %d = %d ", x, y, temp);
        }
        printf_s("\n");
    }
    return 0;
}

void t3()
{
    int x = 0, y = 0, z = 0;
    printf_s("please input 30个字符或者随你吧\n");
    char c = getchar();
    while ((c = getchar()) != '\n')
    {
        if (c >= '0' && c <= '9')
        {
            x++;
        }
        else if (c >= 'a' && c <= 'z' || c >= 'A' && c <= 'Z')
        {
            y++;
        }
        else
        {
            z++;
        }
    }
    printf_s("数字有%d,字母有%d,其余有%d\n", x, y, z);
    return 0;
}

void t2()
{
    int n;
    double sum = 0;
    printf_s("please input n\n");
    scanf_s("%d", &n);
    double a =0;
    int i = 0;
    for (i ; i <= n; i++)
    {    
        a = 1.0 / (i * i + 1);
        if (i % 2 == 0)
        {
            printf_s("%.4lf + ", a);
            sum = sum + a;
        }
        else
        {
            printf_s("%.4lf - ", a);
            sum = sum - a;
        }
    }
    printf_s("0    =%.4lf\n", sum);
    return 0;
}

void t1()
{
    int s = 5;
    int a, b = 0;
    while (s > 0)
    {
        printf_s("选择菜单\n");
        printf_s("        1 求两个数的和\n");
        printf_s("        2 求两个数的差\n");
        printf_s("        3 求两个数的商和余数\n");
        printf_s("        4 求两个数的倒数之和\n");
        printf_s("        0 退出系统\n");
        scanf_s("%d", &s);
        if (s <= 4 && s >= 1)
        {
            printf_s("input 2 number a b\n");
            scanf_s("%d%d", &a, &b);
            if (s == 1)
            {
                printf_s("a+b=%d\n", a + b);
                printf_s("input any key you can return");
                system("pause");
            }
            else if (s == 2)
            {
                printf_s("a-b=%d\n", a - b);
                printf_s("input any key you can return");
                system("pause");
            }
            else if (s == 3)
            {
                printf_s("a/b=%.2lf\n", 1.0 * a / b);
                printf_s("余数为%d\n", a - (a / b) * b);
                printf_s("input any key you can return");
                system("pause");
            }
            else if (s == 4)
            {
                double m = 1.0 / a + 1.0 / b;
                printf_s("1/a+1/b=%.2lf\n", m);
                printf_s("input any key you can return");
                system("pause");
            }
        }
    }
    return 0;
}

int main()
{    
    int s = 0;
    int sm = 1;
    while (sm != 0)
    {
        printf_s("please input the 题目 you want to out put 1-5\n");
        scanf_s("%d",&s);
        if (s == 5)
        {
            t5();
        }
        else if (s == 4)
        {
            t4();
        }
        else if (s == 3)
        {
            t3();
        }
        else if (s == 2)
        {
            t2();
        }
        else if (s == 1)
        {
            t1();
        }
        printf_s("do you 想要再来一次？1/0\n");
        scanf_s("%d", &sm);
    }
    system("pause");
    return 0;
}

运行结果及报错内容

其余的运行都没问题，但是t1和t2都直接跳过了自定义函数

我的解答思路和尝试过的方法

我尝试过把t1和t2的函数单独拿出来测试，都没问题可以运行

在t1和t2函数的开头清空一下缓存，添加如下语句：rewind(stdin); （或者fflush(stdin) ，但是fflush(stdin)在vs2015版本后已经被废弃）。t1和t2函数修改如下：（t3/t4/t5函数最好也在开始清空一下缓存）

void t2()
{
    int n;
    double sum = 0;
    rewind(stdin);
    printf_s("please input n\n");
    scanf_s("%d", &n);
    double a = 0;
    int i = 0;
    for (i; i <= n; i++)
    {
        a = 1.0 / (i * i + 1);
        if (i % 2 == 0)
        {
            printf_s("%.4lf + ", a);
            sum = sum + a;
        }
        else
        {
            printf_s("%.4lf - ", a);
            sum = sum - a;
        }
    }
    printf_s("0    =%.4lf\n", sum);
    return ;
}

void t1()
{
    int s = 5;
    int a, b = 0;
    rewind(stdin);
    while (s > 0)
    {
        printf_s("选择菜单\n");
        printf_s("        1 求两个数的和\n");
        printf_s("        2 求两个数的差\n");
        printf_s("        3 求两个数的商和余数\n");
        printf_s("        4 求两个数的倒数之和\n");
        printf_s("        0 退出系统\n");
        scanf_s("%d", &s);
        if (s <= 4 && s >= 1)
        {
            printf_s("input 2 number a b\n");
            scanf_s("%d%d", &a, &b);
            if (s == 1)
            {
                printf_s("a+b=%d\n", a + b);
                printf_s("input any key you can return");
                system("pause");
            }
            else if (s == 2)
            {
                printf_s("a-b=%d\n", a - b);
                printf_s("input any key you can return");
                system("pause");
            }
            else if (s == 3)
            {
                printf_s("a/b=%.2lf\n", 1.0 * a / b);
                printf_s("余数为%d\n", a - (a / b) * b);
                printf_s("input any key you can return");
                system("pause");
            }
            else if (s == 4)
            {
                double m = 1.0 / a + 1.0 / b;
                printf_s("1/a+1/b=%.2lf\n", m);
                printf_s("input any key you can return");
                system("pause");
            }
        }
    }
    return ;
}