Segment Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description     Silen August does not like to talk with others.She like to find some interesting problems. Today she finds an interesting problem.She…

题目链接: Segment Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others) Problem Description       Silen August does not like to talk with others.She like to find some interesting problems. Today she finds an interesting pro…

题目链接: hdu:http://acm.hdu.edu.cn/showproblem.php?pid=5666 bc(中文):http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=688&pid=1002 题解: 首先可以统计出三角形内所有的点数,(q-2)+(q-3)+...+1=(q-1)*(q-2)/2 其次只要减去被线段割到的点,对于线段xi+yi=q,割到的点有gcd(xi,yi)-1=gcd(xi,q…

Segment Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1082    Accepted Submission(s): 398 Problem Description     Silen August does not like to talk with others.She like to find some interesti…

将斐波那契的前100000个,每个的前40位都插入到字典树里(其他位数删掉),然后直接查询字典树就行. 此题坑点在于 1.字典树的深度不能太大,事实上,超过40在hdu就会MLE…… 2.若大数加法时,保存的位数过少,就会导致低位误差,累积起来就可能导致前40位产生错误,解决办法是提高精度. #include<iostream> #include<cstring> #include<string> #include<cstdio> using namespa…

/** 多校联合2015-muti7-1004 <a target=_blank href="http://acm.hdu.edu.cn/showproblem.php?pid=5372">HDU 5372 Segment Game <span style="font-family: Arial, Helvetica, sans-serif;"></span><span style="font-family: Ar…

C++大数板子 使用样例在主函数里看就好,必要的运算符都重载了. #include <iostream> using namespace std; ;/*精度位数,自行调整*/ //1.如果需要控制输出位数的话,在str()里面把len调成需要的位数 //2.很大的位数是会re的,所以如果是幂运算的话,如 计算x^p的位数n, n=p*log(10)x+1;(注意要加一) //3.还可以加上qmul,取模的过程也就是str(),c_str()再搞一次 class bign{ friend is…

题目:这里 题意:在线段x+y=q与坐标轴围成的三角形中,求有多少个坐标为整数的点,答案模上p. 很容易就想到最后答案就是((q-1)*(q-2))/2然后模上p就是了,但是这个数字比较大,相乘会爆long long,于是用二进制的乘法,类似于快速幂,另外注意这个除以2得在 之前就处理了,因为最后答案一定要是个准确的整数. #include<cstdio> #include<cstring> #include<iostream> #include<algorith…

N! Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 100274    Accepted Submission(s): 30006 Problem Description Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!   Input One N in…

因为刚学fft,想拿这题练练手,结果WA了个爽= =. 总结几点犯的错误: 1.要注意处理前导零的问题. 2.一定要注意数组大小的问题.(前一个fft的题因为没用到b数组,所以b就没管,这里使用了b数组,结果忘记给其大小乘以4倍了) 代码如下: #include<bits/stdc++.h> using namespace std; ; ; typedef long long ll; struct Complex { double x,y; Complex(,) :x(_x),y(_y) {}…