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基础算法

基础算法

作者: 喵个酱 | 来源:发表于2019-03-09 19:12 被阅读0次

一、递推算法

1、顺推算法

兔子的繁殖过程

#include <stdio.h>

#define NUM 13

int main()

{

    int i;

    //初始月份和第一个月的兔子总数为1

    long fib[NUM]={1,1};

    for(i=2;i<NUM;i++)

    {

         //本月的兔子=上两个月兔子之和

        fib[i]=fib[i-1]+fib[i-2];

    }

    for(i=0;i<NUM;i++)

    {

        printf("%d月兔子总数:%d\n",i,fib[i]);

    }

    return 0;

}

2、顺推实例

最后一个月连本带息取出1000

#include <stdio.h>

#define FETCH 1000//最后取出1000

#define RATE 0.0171//利率 /100

int main()

{

    //便于计算

    double corpus[49];

    int i;

    corpus[48]=(double)FETCH;

    for(i=47;i>0;i--)

    {

        corpus[i]=(corpus[i+1]+FETCH)/(1+RATE/12);

    }

    for(i=48;i>0;i--)

    {

        printf("第%d月末本利合计:%.2f\n",i,corpus[i]);

    }

    return 0;

}

二、枚举(穷举)算法

1、算法描述题

               算

----------------

题题题题题

#include <stdio.h>

int main()

{

    int i1,i2,i3,i4,i5;

    long multi,result;

    //算

    for(i1=1;i1<=9;i1++)

    {

        //法

        for(i2=0;i2<=9;i2++)

        {

            //描

            for(i3=0;i3<=9;i3++)

            {

                //述

                for(i4=0;i4<=9;i4++)

                {

                    //题

                    for(i5=0;i5<=9;i5++)

                    {

                        multi=i1*10000+i2*1000+i3*100+i4*10+i5;

                        result=i5*100000+i5*10000+i5*1000+i5*100+i5*10+i5;

                        if(multi*i1==result)

                        {

                            printf("\n%5d%2d%2d%2d%2d\n",i1,i2,i3,i4,i5);

                            printf("X%12d\n",i1);

                            printf("_____________________\n");

                            printf("%3d%2d%2d%2d%2d%2d\n",i5,i5,i5,i5,i5,i5);

                        }

                    }

                }

            }

        }

    }

return 0;

}

2、实例:填运算符

5 5 5 5 5=5

#include <stdio.h>

int main()

{

    int j,i[5];//循环变量,数组i表示4个运算符

    int sign;

    int result;

    int count=0;

    int num[6];

    float left,right;

    char oper[5]={' ','+','-','*','/'};//运算符,第0个元素不用

    printf("请输入5个数:");

    for(j=1;j<=5;j++)

        scanf("%d",&num[j]);

    printf("请输入结果:");

    scanf("%d",&result);

    for(i[1]=1;i[1]<=4;i[1]++)//1:+,2:-,3:*,4:/

    {

        //排除除号后跟0的情况

        if(i[1]<4||num[2]!=0)

        {

        for(i[2]=1;i[2]<=4;i[2]++)

        {

        if(i[2]<4||num[3]!=0)

        {

        for(i[3]=1;i[3]<=4;i[3]++)

        {

        if(i[3]<4||num[4]!=0)

        {

            for(i[4]=1;i[4]<=4;i[4]++)

            {

                if(i[4]<4||num[5]!=0)

                {

                    //左边为0

                    left=0;

                    //右边为第一个数

                    right=num[1];

                    //加法

                    sign=1;

                    for(j=1;j<=4;j++)

                    {

                        //第j个运算符

                       switch(oper[i[j]])

                       {

                            case '+':

                                left=left+sign*right;

                                sign=1;

                                right=num[j+1];

                                break;

                            case '-':

                                left=left+sign*right;

                                sign=-1;

                                right=num[j+1];

                                break;

                            case '*':

                                right=right*num[j+1];

                                break;

                            case '/':

                                right=right/num[j+1];

                                break;

                        }

                    }

                    if(left+sign*right==result)

                    {

                        count++;

                        printf("%3d:",count);

                        for(j=1;j<=4;j++)

                        printf("%d%c",num[j],oper[i[j]]);

                        printf("%d=%d\n",num[5],result);

                    }

                  }

                }

         }

        }

        }

        }

        }

    }

if(count==0)

printf("没有符合要求的方法!\n");

return 0;

}

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