UVa - 12050】的更多相关文章

题目链接:uva 12050 - Palindrome Numbers 题意:求第n个回文串 思路:首先可以知道的是长度为k的回文串个数有9*10^(k-1),那么依次计算,得出n是长度为多少的串,然后就得到是长度为多少的第几个的回文串了,有个细节注意的是, n计算完后要-1! 下面给出AC代码: #include <bits/stdc++.h> typedef long long ll; using namespace std; ; ll num[maxn]; int n,ans[maxn]…
题目大意:给出i,输出第i个镜像数,不能有前导0. 题解:从外层开始模拟 #include <stdio.h> int p(int x) { int sum, i; ;i<=x;i++) sum *= ; return sum; } int main() { ], c; while(~scanf("%d", &n)) { ) break; i=; *p((i-)/)) { n -= *p((i-)/); i++; } c = (i+)/; t=; while(…
A palindrome is a word, number, or phrase that reads the same forwards as backwards. For example,the name “anna” is a palindrome. Numbers can also be palindromes (e.g. 151 or 753357). Additionallynumbers can of course be ordered in size. The first few…
A palindrome is a word, number, or phrase that reads the same forwards as backwards. For example, the name "anna" is a palindrome. Numbers can also be palindromes (e.g. 151 or 753357). Additionally numbers can of course be ordered in size. The f…
A palindrome is a word, number, or phrase that reads the same forwards as backwards. For example,the name “anna” is a palindrome. Numbers can also be palindromes (e.g. 151 or 753357). Additionallynumbers can of course be ordered in size. The first fe…
长度为k的回文串个数有9*10^(k-1) #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include <vector> #include <stack> #include <queue> #include <algorith…
Solve the equation: p ∗ e −x + q ∗ sin(x) + r ∗ cos(x) + s ∗ tan(x) + t ∗ x 2 + u = 0 where 0 ≤ x ≤ 1. Input Input consists of multiple test cases and terminated by an EOF. Each test case consists of 6 integers in a single line: p, q, r, s, t and u (…
A palindrome is a word, number, or phrase that reads the same forwards as backwards. For example, the name “anna” is a palindrome. Numbers can also be palindromes (e.g. 151 or 753357). Additionally numbers can of course be ordered in size. The first…
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UVA - 10564 Paths through the Hourglass 题意: 要求从第一层走到最下面一层,只能往左下或右下走 问有多少条路径之和刚好等于S? 如果有的话,输出字典序最小的路径. f[i][j][k]从下往上到第i层第j个和为k的方案数 上下转移不一样,分开处理 没必要判断走出沙漏 打印方案倒着找下去行了,尽量往左走   沙茶的忘注释掉文件WA好多次   #include <iostream> #include <cstdio> #include <a…