“模拟”类型,题型容易,使用bitset库对二进制逐位操作,初始化.十进制转二进制(unsigned int).位操作. POJ - 3652 Persistent Bits Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u [Submit]   [Go Back]   [Status] Description WhatNext Software creates sequence generators …

主题链接: PKU:http://poj.org/problem?id=3652 ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do? problemId=1933 HDU:http://acm.hdu.edu.cn/showproblem.php?pid=2721 Description WhatNext Software creates sequence generators that they hope will produce fai…

题面: WhatNext Software creates sequence generators that they hope will produce fairly random sequences of 16-bit unsigned integers in the range 0–65535. In general a sequence is specified by integers A, B, C, and S, where 1 ≤ A < 32768, 0 ≤ B < 65536…

简单的贪心和高精度运算,主要还是要读懂题. #include"iostream" #include"stdio.h" #include"string" #include"string.h" #include"cmath" #define mx 5005 using namespace std; ]; char num[mx]; char temp[mx]; bool div(int mod) { ,pre=…

把输入数字每次从9-2除,能整除则记录该数字,最后从小到大输出. 应该算是水题,不过窝第一次写高精度除法,虽然1A,不过中间改了好多次. /****************************************** Problem: 2325 User: Memory: 684K Time: 110MS Language: G++ Result: Accepted ******************************************/ #include <iostrea…

题目有点长,但是题意说的很清楚.位操作. #include <stdio.h> ]; int main() { int a, b, c, s; int i, j, k, n, tmp, mask; while (scanf("%d", &a)!=EOF && a) { scanf("%d%d%d", &b, &c, &s); n = mask = ; nums[n++] = tmp = s; a = a%c…

(-￣▽￣)-* 这道题涉及高精度除法,模板如下: ]; ];//存储进行高精度除法的数据 bool bignum_div(int x) { ,num=; ;s[i];i++) { num=num*+s[i]-'; division[tot++]=num/x+'; num%=x; } division[tot]='\0';//利于进行strcpy() ) //有适合的除数 { ; ') i++; strcpy(s,division+i);//比如49->07,那么下一轮s就变成7,多余的i个0都…

题目传送门 1 2 题意:有向图,所有点先走到x点,在从x点返回,问其中最大的某点最短路程 分析:对图正反都跑一次最短路,开两个数组记录x到其余点的距离,这样就能求出来的最短路以及回去的最短路. POJ 3268 //#include <bits/stdc++.h> #include <cstdio> #include <queue> #include <algorithm> #include <cstring> using namespace…

http://poj.org/problem?id=1743 题意:不可重叠最长重复子串,n<=20000,具体看<后缀数组>-- 罗穗骞 #include <cstdio> #include <algorithm> using namespace std; const int N=20015; void sort(int *x, int *y, int *sa, int n, int m) { static int c[N], i; for(i=0; i<…

// 线段和矩形相交 POJ 1410 // #include <bits/stdc++.h> #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <math.h> using namespace std; #define LL long long typedef p…