两个自然数，计算其和

两个自然数，计算其和，如果输入的答案正确，则显示“right！”，否则提示重做，显示“Not correct! Try again!”;最多给三次机会，如果三次仍未做对，显示“Not correct! You have tried three times! Test over!”,程序结束。

    #include <stdio.h>
    int Add(int a, int b)
    {
        
}
       
    }
    void Print(int flag, int chance)
    {
       if (flag)
          printf("Right!\n");
       else if(chance < 3)
          printf("Not correct. Try again!\n");
       else
          printf("Not correct. You have tried three times!\nTest over!\n");
    }
    int main()
    {
       int  a, b, answer, chance;
       printf("Input a, b:\n");
       scanf("%d,%d", &a, &b);
       chance = 0;
       do
       {
          answer = Add(a, b);
          chance++;
          Print(answer, chance);
       } while ((answer == 0) && (chance < 3));
       return 0;
    }

参考gpt：
根据您提供的代码，我可以帮助您完成这个程序。下面是修改后的代码：

#include <stdio.h>

int Add(int a, int b) {
    return a + b;
}

void Print(int flag, int chance) {
    if (flag) {
        printf("Right!\n");
    } else if (chance < 3) {
        printf("Not correct. Try again!\n");
    } else {
        printf("Not correct. You have tried three times!\nTest over!\n");
    }
}

int main() {
    int a, b, answer, chance;
    printf("Input a, b:\n");
    scanf("%d,%d", &a, &b);
    chance = 0;
    do {
        answer = Add(a, b);
        chance++;
        Print(answer, chance);
    } while ((answer == 0) && (chance < 3));
    return 0;
}

这个程序的作用是获取两个整数a和b的输入，然后计算它们的和。如果输入的答案正确，则显示"Right!"，否则根据尝试次数给出相应的提示。如果尝试三次后仍未得到正确答案，程序显示"Test over!"并结束运行。

请注意，我只修改了代码的格式和语法错误，并保留了原始逻辑。需要您自己完成Add函数的具体实现，即计算两个整数的和并返回结果。

修改如下，改动处见注释，供参考：

#include <stdio.h>
int Add(int a, int b)
{
    return a + b;  //修改
}

void Print(int flag, int chance)
{
    if (flag)
        printf("Right!\n");
    else if (chance < 3)
        printf("Not correct. Try again!\n");
    else
        printf("Not correct. You have tried three times! Test over!\n");
}
int main()
{
    int  a, b, answer, answer1, chance; // 增加一个回答变量 answer1 修改
    printf("Input a, b:\n");
    scanf("%d,%d", &a, &b);
    answer = Add(a, b);     // 修改
    chance = 0;
    do{
        scanf("%d", &answer1);  // 修改
        chance++;
        Print(answer == answer1, chance); //  修改
    } while (chance < 3 && answer != answer1); //while ((answer == 0) && (chance < 3));修改
    return 0;
}