[C++知识库]2021算协训练赛02-小白组题目略解

7-1 有志者，事竟成！

直接输出即可

#include <stdio.h>

int main()
{
    printf("===========================================\n");
    printf(" Nothing is impossible to a willing heart!\n");
    printf("===========================================\n");
    return 0;
}

7-2 3721数

本题考察了if条件判断语句和循环

#include <stdio.h>

int main()
{
    for(int i = 1; i < 100; i++)
    {
        if( i % 3 == 2 && i % 7 == 1 )
            printf("%d\n", i);
    }
    return 0;
}

当然，如果代码能力有限也可以手算，把编译器语言直接改成PHP输入答案即可

8
29
50
71
92

7-3 判断闰年

首先明确什么是闰年？1、能被4整除，但不能被100整除；2、能被400整除； 然后使用if语句判断年份即可
要特别注意输出的内容

#include <stdio.h>

int main()
{
    int year;
    scanf("%d", &year);
    if( year % 400 == 0 || ( year % 4 == 0 && year % 100 != 0))
    {
        printf("YE5");
    }else 
    {
        printf("N0");
    }
    return 0;
}

7-4 365次方

可以使用循环也可以使用math库里面的pow方法

#include <stdio.h>

int main()
{
    double x;
    scanf("%lf", &x);
    printf("%.8f\n%.8f", pow(( 1 + x ), 365 ), pow(( 1 - x ), 365));
    return 0;
}

7-5 简单的鸡兔同笼

7-6 Another chicken and rabbit cage?

鸡兔同笼 略

7-7 求一元二次方程的根

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

int main() {
    double a,b,c,value,value1,complex;
    scanf("%lf %lf %lf",&a,&b,&c);
    double de_ta = b * b - 4 * a * c;
    if (a == 0 && b == 0) {
        if (c == 0) printf("Zero Equation");
        else printf("Not An Equation");
    }else {
        if (de_ta == 0) {
            value = (-1 * b) / (2 * a);
            printf("%.2lf",value);
        }else if (de_ta > 0) {
            // 有特殊情况a = 0时，方程有唯一实数根
            if (a == 0) {
                value = (-1) * (c / b);
                printf("%.2lf",value);
            }else {
                value = (-1 * b - sqrt(de_ta)) / (2 * a);
                value1 = (-1 * b + sqrt(de_ta)) / (2 * a);
                printf("%.2lf\n%.2lf",value1,value);
            }
        }else {
            value1 = sqrt(-1 * de_ta) / (2 * a);
            value = (-1) * (b / (2 * a));
            // 有特殊情况，为纯虚数且前面输出要加上0.00
            if (b == 0) {
                printf("0.00+%.2lfi\n0.00%.2lfi",value1,-1*value1);
            }else {
                // 有两个复数根,先输出虚部是正的
                complex = (value1 > 0) ? value1:(-1)*value1;
                printf("%.2lf+%.2lfi\n%.2lf%.2lfi",value,complex,value,-1*complex);
            }
        }
    }
    return 0;
}

7-8 日K蜡烛图

#include<stdio.h>
int main()
{
	double open=0,close=0,high=0,low=0;
	scanf("%lf %lf %lf %lf",&open,&high,&low,&close);
	
		if(close<open)
		printf("BW-Solid");
		else if(close>open)
		printf("R-Hollow");
		else
		printf("R-Cross"); 
			
	if((low<open&&low<close)&&(high>open&&high>close))
		printf(" with Lower Shadow and Upper Shadow");
	else if(low<open&&low<close)
		printf(" with Lower Shadow");
	else if(high>open&&high>close)
		printf(" with Upper Shadow");
	
	return 0;
}

7-9 Bash博弈

/**
Bash gaame:
(1)二人游戏
(2)n个物品
(3)每个人能够取走1-m个物品
(4)取走最后一个物品的取胜
(5)如果n%(m+1)==0，这个时候先手必败。否则先手必胜。
*/
#include <stdio.h>

int main()
{
	int n[5]={18,23,4,9,15}, m[5]={8,5,9,2,4};
	for (int i = 0; i < 5; i++)
	{
		if (n[i] % (m[i] + 1) == 0)
		{
            printf("GG\n");
		}
		else
		{
            printf("JJ\n");
		}
	}
	return 0;
}