UVa 256 - Quirksome Squares

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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;}

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