PTA攀登者---题集第一期1--10.

落日映苍穹つ 2024-03-31 09:21 20阅读 0赞

![1425ed23b00d44ba8145e5e38d9125d6.png][]

本题就是对于公式的书写，题目比较简单。

#include<stdio.h>
    int main()
    {
        int C;
        C=5*(150-32)/9;
        printf("fahr = 150, celsius = %d\n",C);
        return 0;
    }

![b61b5f78116b4095ad0e2e1b7f296f03.png][]

本题就是对于公式的书写，题目比较简单。

#include <stdio.h>
    int main()
    {
        double s,g = 10,t = 3;
        s = g*t*t/2;
        printf("height = %.2lf\n",s);
        return 0;
    }

![a17f54d1783f422c98f163f1d3e8aba7.png][]

本题就是对于公式的书写，题目比较简单。

#include<stdio.h>
    int main()
    {
        int A,B,C,D,E,F;
        scanf("%d %d",&A,&B);
        C=A+B;
        D=A-B;
        E=A*B;
        F=A/B;
        printf("%d + %d = %d\n",A,B,C);
        printf("%d - %d = %d\n",A,B,D);
        printf("%d * %d = %d\n",A,B,E);
        printf("%d / %d = %d\n",A,B,F);
        return 0;
    }

![b3de01203db344d08db2d7b4c60fa329.png][]

这里考察了if和else的使用，题目比较简单。

#include<stdio.h>
    int main()
    {
        float x;
        scanf("%f",&x);
        if (x == 0)
           printf("f(0.0) = 0.0");
        else
           printf("f(%.1f) = %.1f",x,1 / x);
        return 0;
    }

![12833b6976104db9b89a74e4ef762c81.png][]

#include<stdio.h>
    #include<math.h>
    int main()
    {
        double x,result = 0;
        scanf("%lf",&x);
        if (x >= 0)
    {
           result = sqrt(x);
           printf("f(%.2lf) = %.2lf\n",x,result);
    }
        else
    {
           result = pow(x + 1,2) + 2 * x + 1 / x;
           printf("f(%.2lf) = %.2lf\n",x,result);
    }
        return 0;
    }

这里就是一个简单的套用公式。![817a97a9bcac4353842a0e1bdf275f33.png][]

#include<stdio.h>
    int main()
    {
    	int n;
    	double S,sum=0;
    	scanf("%d",&n);
    	while(n>=1)
    	{
    		S = 1.0/ n;
    		sum = sum + S;
    		n--;
    	}
    	printf("sum = %.6lf\n",sum);
    	return 0;
    }

![619d4a70c49a4ad2b994f8e7e7fd1e58.png][]

一个套用公式的题目，思路链不多，非常简单，套用公式就可以了了。

#include<stdio.h>
    int main()
    {
        int N;
        double q=1,s=0,t=1;
        scanf("%d",&N);
            for(int i=1;i<=N;i++)
            {
                s+=(1/q)*t;
                q+=3;
                t=-t;//每次循环将t取相反数来实现加减交替
            }
        
        printf("sum = %.3lf",s);
        return 0;
        
    }

![558ab0faa23d42b5a2eabefe428e46ee.png][]

这里考察了函数的使用，并不算难，所以希望大家可以掌握这个题目。

#include<stdio.h>
    double fact(int n)
    {
        double sum1 = 1;
        for(int x = 1;x <= n;x++)
        {
            sum1 = sum1 * x;
        }
        return sum1;
    }
    int main()
    {
        int m,n;
        double result;
        scanf("%d %d",&m,&n);
        result = fact(n) / (fact(m) * fact(n-m));
        printf("result = %.0f\n",result);
        return 0;
    }

![f72b9067df564b4b828a9cf8eae01ca4.png][]

#include<stdio.h>
    int main()
    {
        int a,b,c,d;
        scanf("%d %d %d %d",&a,&b,&c,&d);
        printf("Sum = %d; ",a+b+c+d);
        printf("Average = %0.1f",(a+b+c+d)/4.0);
        return 0;
    }

![a1fd1f309cc8401eb5b89d4a73eef2bb.png][]

简单的一个小学应用题，非常基础的一道题目

#include<stdio.h>
    int main()
    {
        double cost,n;
        scanf("%lf",&n);
        if(n < 0)
            printf("Invalid Value!");
        else if(n <= 50)
        {
            cost = n * 0.53;
            printf("cost = %.2lf\n",cost);
        }
        else
        {
            cost = 50 * 0.53 + (n-50)*0.58;
            printf("cost = %.2lf\n",cost);
        }
        return 0;
    }

**总结：本次PTA做的题目比较简单，更加适合初学者，不过有入门就有入土，希望大家学习的时候不论简单与否都要带着认真的态度，这样才能学好。**

本次题目简单了太，不做题解给大家了，当然攀登攀登，开始都是简单的，后面才是难的开始，加油吧

[1425ed23b00d44ba8145e5e38d9125d6.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/7cb8e248a97a46b1817a8f922520a898.png
[b61b5f78116b4095ad0e2e1b7f296f03.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/4cb6397fc27a4f8f825db2c3a52fd652.png
[a17f54d1783f422c98f163f1d3e8aba7.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/0545b6c84dea4494b0ab9ede98c6616f.png
[b3de01203db344d08db2d7b4c60fa329.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/63820a2482db47b19ee94b489d11cf53.png
[12833b6976104db9b89a74e4ef762c81.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/b3e4cee1410b43dda502372b0498d7a5.png
[817a97a9bcac4353842a0e1bdf275f33.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/06b5472319034fc6b46acf9b6ec89d5c.png
[619d4a70c49a4ad2b994f8e7e7fd1e58.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/c21241eb23434c869f30e54f557f7cda.png
[558ab0faa23d42b5a2eabefe428e46ee.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/29988565221e42e5a84cdcd3b005d88c.png
[f72b9067df564b4b828a9cf8eae01ca4.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/33935342af29403db9136ab601eb72e1.png
[a1fd1f309cc8401eb5b89d4a73eef2bb.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/a1fc260a6763478bb4d0e3ba261fe78d.png