C. Game with Strings time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You play the game with your friend. The description of this game is listed below. Your friend creates n distinct strings of the same length m and tells you all ...
题目： CF机子真心强大啊，这样才跑了600ms，给了你n个数的序列，然后m次询问，每次询问求出序列中每个数是 区间[a,b]内的 几个素数的倍数统计一下，然后对于个数求和，看了题目下面的hint很易懂，然后看到a,b的范围有些大哈，2*10^9，不知道怎么处理，但是后来发现，序列中的数 最大为10^7，所以就算a,b,再大也无所谓的，大于序列中的最大数的部分的素数，序列中不会有任何数 是它倍数的，然后就是对10^7以内的 素数进行预处理，...
Given Length and Sum of Digits time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the largest of the numbers that have length m and sum of digits s. The requi...
简单细节题： #include#include#include#include#include#include#define rep(i,a,b) for(int i=(a);i<(b);i++)#define rev(i,a,b) for(int i=(a);i>=(b);i--)#define clr(a,x) memset(a,x,sizeof a)typedef long long LL;using namespace std;const int mod=1e9 +7;const int maxn=105;const int maxm=905;int da[maxn],db[maxn];int main(){ int n,m; while(~scanf("%d%d",&n,&m)) { if(m>9*n||(n!=1&&m==0))...
Palindrome Transformation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to m...
C. Given Length and Sum of Digits... time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the...
C. Given Length and Sum of Digits... time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have a positive integer m and a non-negative int...
看到这题很开心啊，有印象跟以前做过的很像，貌似最近就做过一个，以时间戳为区间来建立树状数组，然后一开始我以为题意是，给x点加val，它以下的所有节点都加-val；所以一开始就以 加 和 减 建立了两个树状数组，最后 减去就是答案，写完发现跟案例对不上啊，读了题目也没发现读错了，对于那句话 我理解错了，后来看了 这个： http://blog.csdn.net/keshuai19940722/article/details/18967661 仔细看看处理部分，我还以为分奇...
C. Hacking Cypher time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Polycarpus participates in a competition for hacking into a new secure messenger. Hes almost won. Having carefully studied the interaction protocol, Polycarpus came to the conclusion that ...
这个题感觉还不错，以前字典树写的是最顺手的，这几次比赛屡屡挂在字典树上也是有阴影了啊~~ 题目大意： 给出一些字符串，对每个字符串进行查询，若没出现过返回OK，若出现过就生成新字符串,格式为原字符串+数，数为这个字符串第几次重复出现。 解题思路： 字典树，对于每个字符串的插入次数进行计数。 下面是代码： #include #include #include #include #include #include #include #include #include...
Vanya and Exams time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vanya wants to pass n exams and get the academic scholarship. He will get the scholarship if the average grade mark for all the exams is at least avg. The exam grade cannot exceed r. Vanya ...
题目: C. Vanya and Exams time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vanya wants to pass n exams and get the academic scholarship. He will get the scholarship if the average grade mark for all the exams is at least avg. The exam grade cannot e...
昨天这场CF打的还挺爽的，不过就是没咋涨Rating，没把握好涨Rating的机会。。 本来可以过四题的，，但是很失败，重评后跪了两道。。唉:-( A. Vasya and Football 思路：给每个人计数，黄牌+1，红牌+2。 当数字第一次超过2时输出。 题目链接：A. Vasya and Football AC代码： #include #include #include #include #include #include using namespace std;struct node { char name[25]; int a[105];}home, a...
不知道到底是我的水平提高了还是CF的题目变水了。。。。。。 A - Vanya and Cubes 水题。。暴力枚举就可以。。 代码如下： #include #include #include #include #include #include #include #include #include #include #include using namespace std;#define LL __int64const int INF=0x3f3f3f3f;int s[100], sum[100];int main(){ int n, i; s[1]=1; sum[i]=1; scanf("%d",&n); for(...
改了时区之后打cf更辛苦了啊。。。昨天没做，今天补了一下啊。 A. Vanya and Cubes 每次加的数规律性很明显就是:(i+1)*i/2。暴力枚举i就可以得到答案。 #include #include #include #include #include #include #include #include #include #include #include #include #define eps 1e-8#define M 1000100#define LL long long//#define LL long long#define INF 0x3f3f3f#define PI 3.1415926535898#define mod 1000000007#d...