http://poj.org/problem?id=2402 #include<iostream> #include<cmath> using namespace std; //int a[20]={9,9,90,90,900,900,9000,9000,90000,90000,900000,900000,9000000,9000000,90000000};//x位上的回文数个数为f(x)=9*10^((x+1)/2); __int64 b[19]={0,9,18,108,198,1098,1998,10998,19998,109998,199998,1099998,1999998, 10999998,19999998,109999998,199999998,1099999998,1999999998};//前x位上的回文数总个数 int c[10]; int main() { __int64 i,j,n,d,k,l; double temp; while(scanf("%I64d",&i)!=EOF&&i) { for(j=0;;j++) { if(i<=b[j]) break; } //j回文数的位数 n=b[j-1];//1...n-1 数的回文数总数 d=i-n-1; temp=(j+1)/2-1; k=(int)pow(10.0,temp); k+=d; l=k; if(j%2==1) l=k/10; for(j=0;l>0;j++) { c[j]=l%10;//前半部分和后半部分对称 l/=10; } printf("%I64d",k);//输出前半部分 for(i=0;i<j;i++)//输出后半部分 printf("%d",c[i]); printf("n"); } return 0; }
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