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  • c++冒泡排序法

    //冒泡排序法
    #include<iostream>
    using namespace std;
    int main (){
        int i,j,t,ii;
        int a[11];        //第0个元素始终没有用
        cout<<"input 10 numbers:"<<endl;
        for(i=1;i<11;i++){
            cin>>a[i];        //输入a[1]~a[10]
        }
        cout<<endl;
        for(j=1;j<=10;j++){        //共进行9趟比较
            for(i=1;i<=11-j;i++){        //在每趟中要进行(10-j)次两两比较
                if(a[i]<a[i+1]){        //如果后面的数大于前面的数
                    t=a[i];
                    a[i]=a[i+1];
                    a[i+1]=t;            //交换两个数的位置,使大数上浮
                    
                }
                cout<<i<<"*";
            }
            cout<<"array:"<<j<<endl;
            cout<<"the sroted numbers:"<<endl;
        for(i=1;i<11;i++){
            cout<<a[i]<<" ";        //输出10个数
        }
        cout<<endl;
        }
    /*    cout<<"the sroted numbers:"<<endl;
        for(i=1;i<11;i++){
            cout<<a[i]<<" ";        //输出10个数
        }
        cout<<endl;
        */
        system("PAUSE");
        return 0;
    }

     运行结果

    input 10 numbers:
    1 2 3 4 5 6 7 8 9 10
    
    1*2*3*4*5*6*7*8*9*10*array:1
    the sroted numbers:
    2 3 4 5 6 7 8 9 10 1
    1*2*3*4*5*6*7*8*9*array:2
    the sroted numbers:
    3 4 5 6 7 8 9 10 2 1
    1*2*3*4*5*6*7*8*array:3
    the sroted numbers:
    4 5 6 7 8 9 10 3 2 1
    1*2*3*4*5*6*7*array:4
    the sroted numbers:
    5 6 7 8 9 10 4 3 2 1
    1*2*3*4*5*6*array:5
    the sroted numbers:
    6 7 8 9 10 5 4 3 2 1
    1*2*3*4*5*array:6
    the sroted numbers:
    7 8 9 10 6 5 4 3 2 1
    1*2*3*4*array:7
    the sroted numbers:
    8 9 10 7 6 5 4 3 2 1
    1*2*3*array:8
    the sroted numbers:
    9 10 8 7 6 5 4 3 2 1
    1*2*array:9
    the sroted numbers:
    10 9 8 7 6 5 4 3 2 1
    1*array:10
    the sroted numbers:
    10 9 8 7 6 5 4 3 2 1
    请按任意键继续. . .
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  • 原文地址:https://www.cnblogs.com/walter371/p/4050479.html
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