UVa 256 - Quirksome Squares

题目：一个偶数位的数字，从中间拆开后的两个数字的和的平方与原来的数字相等。

输入位数，求所有满足上面性质的对应位数的数字。

分析：打表。直接打表计算即可。

说明：写了个生成输出程序的程序。

#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{
    int n;
    while ( cin >> n ) {
        if ( n == 2 ) {
            cout << "00" << endl;
            cout << "01" << endl;
            cout << "81" << endl;
        }
        if ( n == 4 ) {
            cout << "0000" << endl;
            cout << "0001" << endl;
            cout << "2025" << endl;
            cout << "3025" << endl;
            cout << "9801" << endl;
        }
        if ( n == 6 ) {
            cout << "000000" << endl;
            cout << "000001" << endl;
            cout << "088209" << endl;
            cout << "494209" << endl;
            cout << "998001" << endl;
        }
        if ( n == 8 ) {
            cout << "00000000" << endl;
            cout << "00000001" << endl;
            cout << "04941729" << endl;
            cout << "07441984" << endl;
            cout << "24502500" << endl;
            cout << "25502500" << endl;
            cout << "52881984" << endl;
            cout << "60481729" << endl;
            cout << "99980001" << endl;
        }
    }
    return 0;
}

打表程序：

#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{
	printf("#include <iostream>\n");
	printf("#include <cstdlib>\n\n");
	printf("using namespace std;\n\n");
	printf("int main()\n{\n");
	printf("    int n;\n");
	printf("    while ( cin >> n ) {\n");
	printf("        if ( n == 2 ) {\n");
	for ( int i = 0 ; i <= 9 ; ++ i )
	for ( int j = 0 ; j <= 9 ; ++ j )
		if ( (i+j)*(i+j) == (i*10+j) )
			printf("            cout << \"%01d%01d\" << endl;\n",i,j);
	printf("        }\n");
	printf("        if ( n == 4 ) {\n");
	for ( int i = 0 ; i <= 99 ; ++ i )
	for ( int j = 0 ; j <= 99 ; ++ j )
		if ( (i+j)*(i+j) == (i*100+j) )
			printf("            cout << \"%02d%02d\" << endl;\n",i,j);
	printf("        }\n");
	printf("        if ( n == 6 ) {\n");	
	for ( int i = 0 ; i <= 999 ; ++ i )
	for ( int j = 0 ; j <= 999 ; ++ j )
		if ( (i+j)*(i+j) == (i*1000+j) )
			printf("            cout << \"%03d%03d\" << endl;\n",i,j);
	printf("        }\n");
	printf("        if ( n == 8 ) {\n");
	for ( int i = 0 ; i <= 9999 ; ++ i )
	for ( int j = 0 ; j <= 9999 ; ++ j )
		if ( (i+j)*(i+j) == (i*10000+j) )
			printf("            cout << \"%04d%04d\" << endl;\n",i,j);
	printf("        }\n");
	printf("    }\n");
	printf("    return 0;\n}\n");
	return 0;
}