ZOJ 1078 Palindrom Numbers

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ZOJ 1078 Palindrom Numbers

原题链接题目大意:判断一个数是不是palindrom.不限于十进制,可以在任何进制下判断. 解法:还好,数字的范围不大,int类型足够搞定.方法就是从2进制开始,先把数字转换成2进制,判断是否对称,然后转换成3进制,直到16进制.这里的进制转换方法可以留意一下. 参考代码: #include<iostream> using namespace std; bool isPalindrom(int, int); int main(){ int i,n; while(cin>>n&am…

ZOJ Problem Set - 1078 Palindrom Numbers

属于水题,主要是涉及到回文问题. 这里标注下进制转换的方法: while(n) { p[i]=n%basis; n/=basis; } 见代码: #include <stdio.h> int revert(int n,int base) { int p[20]; int i=0; while(n) { p[i++]=n%base; n/=base; } int ispar=1; for(int j=0,k=i-1;j<k;j++,k--) { if(p[j]!=p[k]) { ispar…

zoj 1078

非常简单的题,没什么好说的.唯一值得一提的就是在判断是否是回文的时候只需遍历一半的元素即可,稍微提高一点性能. #include<iostream> using namespace std; int main(void) { int n,i,j,flag; while(cin>>n&&n) { ] = {}; ]; ;i<=;i++) { int m = n; ; while(m) { //获取十进制数化为对应进制数的每一位 c[len++] = m%i; m…

zoj 1383 Binary Numbers

Binary Numbers Time Limit: 2 Seconds Memory Limit: 65536 KB Given a positive integer n, print out the positions of all 1's in its binary representation. The position of the least significant bit is 0. Example The positions of 1's in the binary r…

ZOJ 3365 Integer Numbers

Integer Numbers Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on ZJU. Original ID: 336564-bit integer IO format: %lld Java class name: Main Special Judge The boy likes numbers. He has a sheet of paper. He have written a…

ZOJ 1095 Humble Numbers

原题链接题目大意:定义了一种数字Humble Number,他们的质因数只包含2.3.5.7中的一个或者几个,求第n个这样的数,1<=n<=5842. 解法:一看到这道题又在想DFS了,可是我写的不熟练,最后罢工.Google之,发现还是可以用穷举法解决的.先列出2000000000以内的所有Humble Number,然后再查表.输出的时候要注意下,尾数是1.2.3的和剩下的数字要区别对待,可是尾数是11.12.13又是不一样的. 参考代码: #include<iostream>…

zoj 1828 Fibonacci Numbers

A Fibonacci sequence is calculated by adding the previous two members of the sequence, with the first two members being both 1. f(1) = 1, f(2) = 1, f(n > 2) = f(n - 1) + f(n - 2) Your task is to take a number as input, and print that Fibonacci number…