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. . .

 

posted on 2014-10-25 18:35  walter371  阅读(313)  评论(0编辑  收藏  举报

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