usaco 1.2 Palindromic Squares(进制转化)
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考察进制转化 注意一些细节就可以了
直接上代码:
/*ID: who jayLANG: C++TASK: palsquare*/#include<stdio.h>int x[20],xlen,y[20],ylen,B;void change(int n){ int m; m=n; xlen=0; while(m) { x[++xlen]=m%B; m/=B; } m=n*n; ylen=0; while(m) { y[++ylen]=m%B; m/=B; }}bool ispal(){ bool flag=1; int i,j; for(i=1,j=ylen; i<=j; i++,j--) { if(y[i]!=y[j]) { flag=0; return 0; break; } } if(flag) return 1;}int main(){ FILE *fin = fopen ("palsquare.in", "r"); FILE *fout = fopen ("palsquare.out", "w"); char c[]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J'}; int i,j; fscanf(fin,"%d",&B); for(i=1; i<301 ;i++) { change(i); if(ispal()) { for(j=xlen; j>0 ;j--) fprintf(fout,"%c",c[x[j]]); fprintf(fout," "); for(j=ylen; j>0 ;j--) fprintf(fout,"%c",c[y[j]]); fprintf(fout,"\n"); } } return 0;}
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