这道题目最开始完全不懂,后来百度了一下,原来是字典序.而且还是组合数学里的东西.看字典序的算法看了半天才搞清楚,自己仔细想了想,确实也是那么回事儿.对于长度为n的数组a,算法如下:(1)从右向左扫描,找到满足a[i]<a[i+1]的第一个i,也就是i = max{i|a[i]<a[i+1]},同时也意味着a[i+1]~a[n]是升序:(2)从右向左扫描,找到满足a[j]>a[i]的第一个j,也就是j = max{j|a[j]>a[i]},a[j]也是满足大于a[i]的最小数:(3)…

这道题搞了很久啊.搜索非常好的一道题.昨天想了2小时,以为是深搜,但后来发现深搜怎么也没法输出正确路径.今天拿宽搜试了一下,问题就是普通的队列宽搜没法得到当前时间最小值.看了一下讨论区,发现优先级队列.好久不用了,都忘记了.各种忘记,优先级队列排序都忘掉了.搞了好半天.最后还需要注意的是格式化输出,采用栈格式输出.需要保存每个节点的移动方向.并且注意若终点是怪兽,还是需要"Fight".这道题目感觉不是一道水题,还挺不错. #include <iostream> #incl…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1027 Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10388    Accepted Submission(s): 5978 Problem Description Now our…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4865    Accepted Submission(s): 2929 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4865    Accepted Submission(s): 2929 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9380    Accepted Submission(s): 5481 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9458    Accepted Submission(s): 5532 Problem Description Now our hero finds the door to the BEelzebub feng5166. He op…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12948    Accepted Submission(s): 7412 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

数学归纳法,得证只需求得使18+ka被64整除的a.且a不超过65. #include <stdio.h> int main() { int i, j, k; while (scanf("%d", &k) != EOF) { j = ; ; i<; i++) { +k*i) % == ) { j = ; break; } } if (j) printf("%d\n", i); else printf("no\n"); }…

大意是给你1个整数n,问你能拆成多少种正整数组合.比如4有5种: 4 = 4;  4 = 3 + 1;  4 = 2 + 2;  4 = 2 + 1 + 1;  4 = 1 + 1 + 1 + 1; 然后就是母函数模板题……小于n的正整数每种都有无限多个可以取用. (1+x+x^2+...)(1+x^2+x^4+...)...(1+x^n+...) 答案就是x^n的系数. #include<cstdio> #include<cstring> using namespace std;…