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加法问题 不同字母加法竖式

xianyun1994 发布于 2010-02-05 19:33, 1446 次点击
给定等式  A B C D E     其中每个字母代表一个数字,且不同数字对应不
                D F G     同字母。编程求出这些数字并且打出这个数字的
             +  D F G     算术计算竖式。

             ───────

                X Y Z D E
忘了在那里复制的 救下

程序代码:
#include<iostream>
using namespace std;

int main()
{
    int abcde,dfg,xyzde;
    for (int a=1;a<10;a++){
        for (int b=0;b<10;b++){
            for (int c=0;c<10;c++){
                for (int d=1;d<10;d++){
                    for (int e=0;e<10;e++){
                        for (int f=0;f<10;f++){
                            for (int g=0;g<10;g++){
                                for (int x=1;x<10;x++) {
                                    for (int y=0;y<10;y++) {
                                        for (int z=0;z<10;z++) {
                                            abcde=a*100000+b*10000+c*1000+d*100+e*10;
                                            dfg=2*(d*100+f*10+g);
                                            xyzde=x*10000+y*1000+z*100+d*10+e;
                                            if (xyzde==abcde+dfg){
                                                cout<<" "<<a<<b<<c<<d<<e<<endl;
                                                cout<<"    "<<d<<f<<g<<endl;
                                                cout<<"+"<<"   "<<d<<f<<g<<endl;
                                                cout<<endl;
                                                cout<<"───────"<<endl<<endl;
                                                cout<<" "<<x<<y<<z<<d<<e<<endl<<endl<<endl;
                                                break;
                                            }}}}}}}}}}}
    cout<<"结束!"<<endl;
    int q;
    cin>>q;
    return 0;
}
别笑我
4 回复
#2
pangding2010-02-06 00:06
我并不是想笑,只是我忽然发现你的代码像架飞机~~
#3
mghxz2522010-02-10 10:05
本来不觉得,经二楼的一说,真有点像飞机。~~~~~~~~····
#4
木瓜君2010-02-10 22:45
时间复杂度~~
#5
shiyuehai2010-02-17 19:46
那么多循环......看都不想看了.....
1