http://acm.hdu.edu.cn/showproblem.php?pid=1573 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6718    Accepted Submission(s): 2342 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X m…

链接:pid=1573">http://acm.hdu.edu.cn/showproblem.php? pid=1573 题意:求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], -, X mod a[i] = b[i], - (0 < a[i] <= 10). 思路:中国剩余定理的模板题(全部除数相互不互质版),假设找不到这种数或者最小的X大于N.输出零. 资料:http:/…

CRT模板题 /** @Date : 2017-09-15 13:52:21 * @FileName: HDU 1573 CRT EXGCD.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$ */ #include <bits/stdc++.h> #define LL long long #define PII p…

HDU 1573 X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4857    Accepted Submission(s): 1611 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2],…

X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2980    Accepted Submission(s): 942 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod…

数论题,本想用中国剩余定理,可是取模的数之间不一定互质,用不了,看到网上有篇文章写得很好的:数论——中国剩余定理(互质与非互质),主要是采用合并方程的思想: 大致理解并参考他的代码后便去试试hdu上这道题,可还是wa了数遍. #include<cstdio> #define scd(x) scanf("%d",&x) #define sclld(x) scanf("%I64d",&x) #define prd(x) printf(&quo…

Problem Description 要求(A/B)%9973,但由于A很大,我们只给出n(n=A%9973)(我们给定的A必能被B整除,且gcd(B,9973)= 1). Input 数据的第一行是一个T,表示有T组数据. 每组数据有两个数n(0 <= n < 9973)和B(1<= B <= 10^9). Output 对应每组数据输出(A/B)%9973. Sample Input 2 1000 53 87 123456789 Sample Output 7922 6060…

X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8365    Accepted Submission(s): 3037 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod…

2891 -- Strange Way to Express Integers import java.math.BigInteger; import java.util.Scanner; public class Main { static final BigInteger ZERO = new BigInteger("0"); static final BigInteger ONE = new BigInteger("1"); static BigInteger…

X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5221    Accepted Submission(s): 1761 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod…

题目链接 题意 : 中文题不详述. 思路 : 中国剩余定理.求中国剩余定理中解的个数.看这里看这里 #include <stdio.h> #include <iostream> #include <math.h> using namespace std ; long long x,y ; long long N,M ; long long ext_gcd(long long a,long long b) { long long t,d ; ) { x = ; y = ;…

只是套模板而已(模板其实也不懂). 留着以后好好学的时候再改吧. 题意—— X = a[i] MOD b[i]; 已知a[i],b[i],求在[1, n]中存在多少x满足条件. 输入—— 第一行一个整数t,表示一共t组数据. 第二行两个整数n,m,表示在n以内寻找满足的数,一共m组方程组. 输出—— 一个整数.如果存在满足的x,则输出x的数量.否则输出0. 直接给代码吧—— #include <cstdio> #include <iostream> #include <cma…

X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8354    Accepted Submission(s): 3031 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], -, X mo…

/* 同余方程组为 X = ri (mod ai) 在范围内求X的个数 先求出特解 X0: 求出 ai数组的LCM: 则有 Xi = X0+LCM 均能满足方程组,判断是否在范围内!! */ #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; typedef long long…

X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod a[i] = b[i], … (0 < a[i] <= 10).   Input 输入数据的第一行为一个正整数…

又来一发水题. 解同余方程而已,用类似于剩余定理的方法就O了. 直接上代码:(注意要判断是否有解这种情况) #include <iostream> #include <cstdio> #define ll long long using namespace std; ll c[],m[],n,t,tot; void exgcd(ll A,ll B,ll& d,ll& x,ll& y) { ) { x=,y=,d=A; } else { exgcd(B,A%B…

X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4869    Accepted Submission(s): 1617 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod…

扩展中国剩余定理的板子,合并完之后算一下范围内能取几个值即可(记得去掉0) #include<iostream> #include<cstdio> #include<cmath> using namespace std; const int N=15; int T,n,m; long long a[N],b[N],A,B,x,y,d; bool fl; void exgcd(long long a,long long b,long long &d,long lo…

题目意思很直接就是一道裸的解线性同余模方程组的题目 #include <cstdio> #include <cstring> using namespace std; #define N 15 int r[N] , m[N]; int ex_gcd(int a , int &x , int b , int &y) { ){ x = , y = ; return a; } int ans = ex_gcd(b , x , a%b , y); int t = x; x…

题目:求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], -, X mod a[i] = b[i], - (0 < a[i] <= 10). 解法:先同上题一样用拓展欧几里德求出同余方程组的最后一个方程 X=ax+b,再调整 x 来求得 X 的解的个数.一些解释请看下面的代码. 注意--每次联立方程后求最小正整数解,可以提高代码速度. 1 #include<cstdio> 2 #i…

HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201…

Leftmost Digit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12229    Accepted Submission(s): 4674题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1060 Problem Description Given a positive integ…

A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4383    Accepted Submission(s): 1573 Problem Description Jimmy experiences a lot of stress at work these days, especiall…

转载来自:http://www.cppblog.com/acronix/archive/2010/09/24/127536.aspx 分类一: 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029.1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093.1094.1095.1096.1097.1098.1106.1108.1157…

模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201 120…

http://acm.hdu.edu.cn/showproblem.php?pid=1573 X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4439    Accepted Submission(s): 1435 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0],…

转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029. 1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093. 1094.1095.1096.1097.1098.1106.1108.1157.116…

A/B Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7310    Accepted Submission(s): 5798 Problem Description 要求(A/B)%9973,但由于A很大,我们只给出n(n=A%9973)(我们给定的A必能被B整除,且gcd(B,9973) = 1).   Input 数据的第一行是一…

HDU分类 http://www.cnblogs.com/ACMan/archive/2012/05/26/2519550.html#2667329 努力A完.方便自己系统A题 不断更新中.................. 水题:1001 1004 简单题1005 找规律 (循环点,周期问题)1008 1012 1013 简单题(有个小陷阱,大数)1017 1018 简单数学题 1019 简单数学题 1020 简单的字符串处理 1021 找规律的数学题,周期81030 简单题,找规律的数学题1…

Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194    Accepted Submission(s): 3345 Problem Description 话说上回讲到海东集团面临内外交困,公司的元老也只剩下XHD夫妇二人了.显然,作为多年拼搏的商人,XHD不会坐以待毙的.  一天,当他正在苦思冥想解困良策的…