C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19536   Accepted: 5204 Description A Compiler Mystery: We are given a C-language style for loop of type for (variable = A; variable != B; variable += C) statement; I.e., a loop w…

题意:有一个在k位无符号整数下的模型:for (variable = A; variable != B; variable += C)  statement; 问循环的次数,若"永不停息"(←_←)*,就输出"FOREVER". 解法:用拓展欧几里德方法求出gcd最大公因数,再利用同余性质转化,求同余方程,或者不定方程.其中题目可化为 a+cx=b(mod 2^k) → cx=b-a(mod 2^k),求最小正整数解.也是求解同余方程. 先将方程化为一般形式:ax=…

POJ 2115:http://poj.org/problem?id=2115 思路 设循环T次 则要满足A≡(B+CT)(mod 2k) 可得 A=B+CT+m*2k 移项得C*T+2k*m=B-A (因为要满足B大于A)即是Exgcd的标准式子了 代码 #include<iostream> #include<cstdio> using namespace std; #define ll long long ll A,B,C,T,k; int gcd(ll a,ll b) { i…

题目地址:POJ 2115 水题. . 公式非常好推.最直接的公式就是a+n*c==b+m*2^k.然后能够变形为模线性方程的样子,就是 n*c+m*2^k==b-a.即求n*c==(b-a)mod(2^k)的最小解.(真搞不懂为什么训练的时候好多人把青蛙的约会都给做出来了,这题却一直做不出来.. . . . 这两道不都是推公式然后变形吗. .... ) 代码例如以下: #include <iostream> #include <cstdio> #include <strin…

http://poj.org/problem?id=3321 http://acm.hdu.edu.cn/showproblem.php?pid=3887 POJ 3321: 题意:给出一棵根节点为1的边不一定的树,然后给出问题:询问区间和 或者 节点值更新. HDU 3887: 题意:和POJ 3321的题意差不多,只不过对每个节点询问不包含该节点的区间和 思路:今天才学了下才知道有DFS序这种东西,加上树状数组处理一下区间和 和 节点更新. DFS序大概就是我们在DFS遍历一棵树的时候,在进…

POJ 3252:Round Numbers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10099 Accepted: 3669 Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Sciss…

http://poj.org/problem?id=2115 题意: 给你一个变量,变量初始值a,终止值b,每循环一遍加c,问一共循环几遍终止,结果mod2^k.如果无法终止则输出FOREVER. 思路: 根据题意原题可化成c * x = b - a mod (2 ^ k),然后解这个模线性方程. #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #includ…

题目:http://poj.org/problem?id=2115 exgcd裸题.注意最后各种%b.注意打出正确的exgcd板子.就是别忘了/=g. #include<iostream> #include<cstdio> #include<cstring> #define ll long long using namespace std; ll a,b,x,y,r,A,B,C,k,g; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}…

C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 23637   Accepted: 6528 Description A Compiler Mystery: We are given a C-language style for loop of type for (variable = A; variable != B; variable += C) statement; I.e., a loop w…

C Looooops Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22704 Accepted: 6251 Description A Compiler Mystery: We are given a C-language style for loop of type for (variable = A; variable != B; variable += C) statement; I.e., a loop which…