C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 24355   Accepted: 6788 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…

扩展GCD...一定要(1L<<k),不然k=31是会出错的 ....                        C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15444   Accepted: 3941 Description A Compiler Mystery: We are given a C-language style for loop of type for (variable…

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: 20128 Accepted: 5405 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…

C Looooops DescriptionA 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 starts by setting variable to value A and while variable is not equal to B, repea…

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…

C Looooops Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 10   Accepted Submission(s) : 3 Problem Description A Compiler Mystery: We are given a C-language style for loop of type for (variable…

题目链接:http://poj.org/problem?id=2115 C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22912   Accepted: 6293 Description A Compiler Mystery: We are given a C-language style for loop of type for (variable = A; variable != B; vari…

poj2115 C Looooops 题意: 对于C的for(i=A ; i!=B ;i +=C)循环语句,问在k位存储系统中循环几次才会结束. 若在有限次内结束,则输出循环次数. 否则输出死循环. (k位==mod $2^{k}$) 列出方程:$A+Cx\equiv B(mode\quad 2^{k})$ 转换一下:$Cx+ky=B-A$ 用exgcd解出 $Cx+ky=gcd(C,k)$ 然后把求出的$x*(B-A)/gcd(C,k)$ 再$\% (k/gcd(C,k))$求个最小正整数解…

C Looooops Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2115 Description A Compiler Mystery: We are given a C-language style for loop of type for (variable = A; variable != B; variable += C) statemen…

1640:C Looooops 时间限制: 1000 ms         内存限制: 524288 KB [题目描述] 原题来自:CTU Open 2004 对于 C 语言的 for (variable = A; variable != B; variable += C)  statement; 循环语句,问在 k 位存储系统中循环几次才会结束.若在有限次内结束,则输出循环次数.否则输出死循环. [输入] 多组数据,每组数据一行四个整数 A,B,C,k.k 表示 k 位存储系统. 读入以0 0…

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…

C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 29262   Accepted: 8441 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: 23616   Accepted: 6517 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…

题目链接:http://poj.org/problem?id=2115 C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 27838   Accepted: 7930 Description A Compiler Mystery: We are given a C-language style for loop of type for (variable = A; variable != B; vari…

C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 29061   Accepted: 8360 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…

看了半天的同余 扩展欧几里得 练练手 C Looooops Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 27079   Accepted: 7690 Description A Compiler Mystery: We are given a C-language style for loop of type for (variable = A; variable != B; variable += C) statem…

Looooops(点击) 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 starts by setting variable to value A and while variable is not equal to B, repeats state…

题意不难理解,看了后就能得出下列式子: (A+C*x-B)mod(2^k)=0 即(C*x)mod(2^k)=(B-A)mod(2^k) 利用模线性方程(线性同余方程)即可求解 模板直达车 #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef long l…

本题和poj1061青蛙问题同属一类,都运用到扩展欧几里德算法,可以参考poj1061,解题思路步骤基本都一样.一,题意: 对于for(i=A ; i!=B ;i+=C)循环语句,问在k位存储系统中循环几次才会结束. 比如:当k=4时,存储的数 i 在0-15之间循环.(本题默认为无符号) 若在有限次内结束,则输出循环次数. 否则输出死循环.二,思路: 本题利用扩展欧几里德算法求线性同余方程,设循环次数为 x ,则解方程 (A + C*x) % 2^k = B ;求出最小正整数 x. 1,化简方…

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 starts by setting variable to value A and while variable is not equal to B, repeats statem…

无符号k位数溢出就相当于mod 2k,然后设循环x次A等于B,就可以列出方程: $$ Cx+A \equiv B \pmod {2^k} $$ $$ Cx \equiv B-A \pmod {2^k} $$ 最后就用扩展欧几里得算法求出这个线性同余方程的最小非负整数解. #include<cstdio> #include<cstring> #define mod(x,y) (((x)%(y)+(y))%(y)) #define ll long long ll exgcd(ll a,…

http://poj.org/problem?id=2115 题解:一个变量从A开始加到B,每次加C并mod2^k,问加多少次.转化为不定方程:C*x+2^K*Y=B-A //poj2115 #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> using namespace std; typedef long long LL; LL bit[]; LL tx,ty;…

题目链接 这个题犯了两个小错误,感觉没错,结果怒交了20+遍,各种改看别人题解,感觉思路没有错误,就是wa. 后来看diccuss和自己查错,发现自己的ecgcd里的x*(a/b)写成了x*a/b.还有(LL)1<<k 写成了 (LL)(1<<k),记住了... 题意: 对于C的for(i=A ; i!=B ;i +=C)循环语句,问在k位存储系统中循环几次才会结束. 若在有限次内结束,则输出循环次数. 否则输出死循环.取最小的满足 cx mod (2^k) = b - a的正x.…

题意:很明显,我就不说了 分析:令n=2^k,因为A,B,C<n,所以取模以后不会变化,所以就是求(A+x*C)%n=B 转化一下就是求 C*x=B-A(%n),最小的x 令a=C,b=B-A 原式等于ax=b(mod n) 这就是标准的解模线性方程 该方程有解的充要条件是d=gcd(n,a) && d|b(可以根据这一条判断是否FOREVER) 然后参考算法导论应用扩展欧几里得求解x a*x0+n*y0=d x=x0*(b/d)(mod n) 然后应用多解条件求最小正整数解,即解的…

题目大意 求同余方程Cx≡B-A(2^k)的最小正整数解 题解 可以转化为Cx-(2^k)y=B-A,然后用扩展欧几里得解出即可... 代码: #include <iostream> using namespace std; typedef long long LL; void extended_gcd(LL a,LL b,LL &d,LL &x,LL &y) { if(!b) { d=a,x=,y=; } else { extended_gcd(b,a%b,d,y,x…

辗转相除法(欧几里得算法) 时间复杂度:在O(logmax(a, b))以内 int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } 扩展欧几里得算法 时间复杂度和欧几里得算法相同 int extgcd(int a, int b, int& x, int& y) { int d = a; if (b != 0) { d = extgcd(b, a % b, y, x); y -= (a / b) * x;…

http://poj.org/problem?id=2115 题意:对于C的循环(for i = A; i != B; i+=C)问在k位存储系统内循环多少次结束: 若循环有限次能结束输出次数,否则输出 FOREVER: 解:设x为循环次数:  (A+C*x)%2^k = B; 则 C*x+A = 2^k*y+B; 所以 C*x - 2^k*y = B-A; 类似于a*x+b*y = c (或 a*x = c(mod b))模线性方程的形式,所以可以根据扩展欧几里得算法解决 #include<s…

题目链接. 分析: 数论了解的还不算太多,解的时候,碰到了不小的麻烦. 设答案为x,n = (1<<k), 则 (A+C*x) % n == B 即 (A+C*x) ≡ B (mod n) 化简得 C*x ≡ (B-A) (mod n) 设 a = C, b = (B-A) 则原式变为 ax=b.解 x. 到这里,以为求出来 a 的逆, 然后 x = b*a-1. a 的 逆好求,用<训练指南>上的模板(P122) inv函数. 例如,求 2 模 66536 下的逆, inv(2,…

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 starts by setting variable to value A and <= x < 2k) modulo 2k. Input The input consists…