国产精品s色,久久黄色影院,欧美主播一区二区三区

SGU 118 Digital Root

Brian — Thu, 19 Aug 2010 07:35:00 GMT

f(n) 是正整数n 的各位数字之和。如�?f(n) �?位数那么n的数根就是f(n)�Q�否则f(n) 的数根就是n 的数栏V��D例说明：987的数�Ҏ��6 (9+8+7=24 2+4=6)。你的�Q务是扑և��q�样表达式的数根: A1*A2*…*AN + A1*A2*…*AN-1 + … + A1*A2 + A1�?

输入包含K个测试样例。输入第一行会�l�出 K (1<=K<=5)。每个测试样例占一行。这一行的�W�一个数是一个正整数N (N<=1000)�?接着是N个非负整�?(序列 A)�?均不��过10⁹�?/p> 数根(癑ֺ�知道): http://zhidao.baidu.com/question/29789035.html
对于求余数的问题是我一开始就忽略了的地方�Q�后来看了HPF的博客，才知道有�q�样的技�?
(A+B)mod C = (A mod C) + (B mod C)
(AB) mod C = (A mod C) × (B mod C)

C�~�译�?, 0MS 0KB , 如果你把前两句定义变量的语句对调�Q�将会耗费 20MS 的时�?br>

#include <stdio.h>
int main()
{
    int i=0,j,K,N,temp,m;
    long int A;
    scanf("%d",&K);
    for (; i<K; i++) {
        scanf("%d",&N);
        m=1;
        A=0;
        for (j=0; j<N; j++) {
            scanf("%d",&temp);
            temp%=9;
            m=(m*temp)%9;
            A=(A+m)%9;
        }
        printf("%d\n",(A+8)%9+1);
    }
    return 0;
}

Brian 2010-08-19 15:35 发表评论

SGU 111 Very simple problem

Brian — Thu, 19 Aug 2010 06:36:00 GMT

下面的代码没有AC�Q�没有AC的原因是 PE on test 8, 注意�Q�是�W�八�l�。我难以理解�Q?span style="BACKGROUND-COLOR: yellow; COLOR: red">希望大牛们指�?/span>�Q?br>

#include <iostream>
#include <stdio.h>
using namespace std;

int main()
{
    __int64 x,left,right,mid;
    scanf("%I64d",&x);
    left=0;
    right=x+1;
    while ((left+1)<right) // 二分查找
    {
        mid=(left+right)/2;
        if(mid*mid<=x)
          left=mid;
        else
          right=mid;
    }
    printf("%I64d",left);
    return 0;
}

Brian 2010-08-19 14:36 发表评论

SGU 107 987654321 problem

Brian — Tue, 17 Aug 2010 05:27:00 GMT

For given number N you must output amount of N-digit numbers, such, that last digits of their square is equal to 987654321.

Input

Input contains integer number N (1<=N<=106)
首先惛_��的是枚�D�Q�我��实也这么做了，5~8无果�Q�对�?�Q�我�?5�q�方�?Pentium M 上跑了��?分半钟之久，我一开始还以�ؓ��d�@环了�Q�后来出来八个结果，也去�|�上核对了一下，��实满��的只有这八个:
   111111111
   119357639
   380642361
   388888889
   611111111
   619357639
   880642361
   888888889
我没有学�q�数论，扑ֈ�一��博文，对于数论解法讲的�q�算清楚�Q�不�q�用pascal写就�Q�我自己用C写了一下，�W�三�ơAC了。博文链接如下：http://blog.csdn.net/Skyprophet/archive/2009/10/05/4634801.aspx

#include <stdio.h>
int main()
{
    int N,i=0;
    scanf("%d",&N);
    if (N<=8)
        printf("0\n");
    else if (N==9)
        printf("8\n");
    else
    {
        printf("72");
        for (; i<N-10; i++)
            printf("0");
        printf("\n");
    }
    return 0;
}

336MS 0K

Brian 2010-08-17 13:27 发表评论

SGU 105 Div 3

Brian — Tue, 17 Aug 2010 05:26:00 GMT

There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Given first N elements of that sequence. You must determine amount of numbers in it that are divisible by 3. (扑և��q�个数列1-N号元素中能被3整除的有多少�?

Input

Input contains N (1<=N<=2³¹ - 1).

一个数乘以10以后�Q�模3后结果不变，�W?�Q?�Q?�Q?1�Q?4�Q?7........��Ҏ��?br>

#include <stdio.h>
int main()
{
    int N;
    scanf("%d",&N);
    N--;
    printf("%d\n", N-(N/3));
    return 0;
}  // 0K  0MS

Brian 2010-08-17 13:26 发表评论

SGU 102 Coprimes

Brian — Tue, 17 Aug 2010 05:24:00 GMT

题目��单翻译：对于�l�出的整�?N (1<=N<=10⁴) �Q�找出所有不大于N且与N互质的正整数个数�?br>两个正整数A和B互质的意思是它们的最大公�U�数�?。现看一下百度百�U�上关于质数的定义吧�?br>当我采用直接数的办法AC后，我发�?font color=#3366ff>跟质数完全没有关�p�R�?/font>

#include<stdio.h>

int gcd(int m,int n) // greatest common divisor
{
    int r=m%n;
    while(r)         // 辗�{盔R��法，TAOCP的第一个算�?/span>
    {
        m=n;
        n=r;
        r=m%n;
    }
    return n;
}

int main()
{
    int N,c=0,i=1; // must start with 1 !
    scanf("%d",&N);
    for (; i<=N; i++)
        if (gcd(N,i)==1) // question hint
            c++;
    printf("%d",c);
    return 0;
}
//  102  .C Accepted 0 ms 0 kb

Brian 2010-08-17 13:24 发表评论

SGU 113 Nearly prime numbers

Brian — Tue, 17 Aug 2010 05:23:00 GMT

Nearly prime number is an integer positive number for which it is possible to find such primes P₁ and P₂ that given number is equal to P₁*P₂. There is given a sequence on N integer positive numbers, you are to write a program that prints “Yes” if given number is nearly prime and “No” otherwise.

Input

Input file consists of N+1 numbers. First is positive integer N (1£N£10). Next N numbers followed by N. Each number is not greater than 10⁹. All numbers separated by whitespace(s).

Output

Write a line in output file for each number of given sequence. Write “Yes” in it if given number is nearly prime and “No” in other case.
不用��Nearly prime numbers 到底是个什么数�Q��M��是两个质数的乘积��对了。枚丄��范围: 2 ~ 32000 (10^4.5)
我的思�\�? 首先�?~32000之间的所有素数都存放在一个数�l�里�Q�然后当你输入一个数据时�Q�先让其逐一模除�q�个数组里的素数�Q�一旦模除结果�ؓ0�Q�则计算��Z��们的商，再判断商是否也�ؓ素数�?strong>注意数据是两个数的��^方的处理

#include <stdio.h>
#include <math.h>
int prime[30000],M=0;

int isP(int n) //判断是否为素�?非常�_��而高�?/span>
{
    int i=2,t=sqrt(n);
    if ((n != 2 && !(n % 2)) || (n != 3 && !(n % 3)) ||
        (n != 5 && !(n % 5)) || (n != 7 && !(n % 7)))
        return 0;
    for (; i<=t; i++)
        if (n%i == 0)
            return 0;
    return 1;
}

int isNP(int n)
{
    int i=0,t=sqrt(n),r;
    for (; prime[i]<=t; i++) // �q�x��在此处理
    {
        if (n%prime[i] == 0) // 模除试商
        {
            r=n/prime[i]; // 求商
            if (isP(r))
                return 1;
        }
    }
    return 0;
}

int main()
{
    int i=3,N,m;
    for (prime[M++]=2; i<32000; i+=2)
        if (isP(i))
            prime[M++]=i;

    scanf("%d",&N);
    while (N--)
    {
        scanf("%d",&m);
        if (m==6)
            printf("Yes\n"); // �E�序中唯一未解决的问题,望各路大牛指�?
        else printf("%s\n",isNP(m) ? "Yes":"No");
    }
    return 0;
}

Brian 2010-08-17 13:23 发表评论

SGU 112 a^b-b^a (Java Edition)

Brian — Tue, 17 Aug 2010 05:21:00 GMT

You are given natural numbers a and b. Find a^b-b^a.

Input

Input contains numbers a and b (1≤a,b≤100).

Output

Write answer to output.

Sample Input

2 3

Sample Output

-1

一看到�q�种题目�Q�就惛_��高精度算法，Java的大数�?br>此�D��实��氓了一点，不过��实�q�了�Q�鄙人的�W�一题SGU啊，不许打击�?br>�q�段旉��看看能不能写�?(C++ Edition) ,啥也别说了，上代码�?br>

import java.math.*;
import java.util.*;

public class Solution
{
   public static void main(String[] args)
   {
      Scanner in = new Scanner(System.in);
      int a = in.nextInt();
      int b = in.nextInt();

      BigInteger A=BigInteger.valueOf(a);
      BigInteger B=BigInteger.valueOf(b); // A^B
      BigInteger rA=BigInteger.valueOf(a);
      BigInteger rB=BigInteger.valueOf(b); // store the result after computing

      for(int i=1; i<b; i++) // just b-1 times
          rA=rA.multiply(A);
      for(int i=1; i<a; i++)
          rB=rB.multiply(B);
      System.out.println(rA.subtract(rB)); // sub
   }
}

附《Core Java I》里关于大数的简单介�l�，讲的��比较清晰的了：
Big Numbers
If the precision of the basic integer and floating-point types is not sufficient, you can
turn to a couple of handy classes in the java.math package: BigInteger and BigDecimal. These
are classes for manipulating numbers with an arbitrarily long sequence of digits. The
BigInteger class implements arbitrary precision integer arithmetic, and BigDecimal does the
same for floating-point numbers.
Use the static valueOf method to turn an ordinary number into a big number:
BigInteger a = BigInteger.valueOf(100);
Unfortunately, you cannot use the familiar mathematical operators such as + and * to
combine big numbers. Instead, you must use methods such as add and multiply in the big
number classes.
BigInteger c = a.add(b); // c = a + b
BigInteger d = c.multiply(b.add(BigInteger.valueOf(2))); // d = c * (b + 2)
C++ NOTE: Unlike C++, Java has no programmable operator overloading. There was no way
for the programmer of the BigInteger class to redefine the + and * operators to give the add and
multiply operations of the BigInteger classes. The language designers did overload the + operator
to denote concatenation of strings. They chose not to overload other operators, and they
did not give Java programmers the opportunity to overload operators in their own classes.
Listing 3–6 shows a modification of the lottery odds program of Listing 3–5, updated to
work with big numbers. For example, if you are invited to participate in a lottery in
which you need to pick 60 numbers out of a possible 490 numbers, then this program
will tell you that your odds are 1 in 7163958434619955574151162225400929334117176
12789263493493351 013459481104668848. Good luck!
The program in Listing 3–5 computed the statement
lotteryOdds = lotteryOdds * (n - i + 1) / i;
When big numbers are used, the equivalent statement becomes
lotteryOdds = lotteryOdds.multiply(BigInteger.valueOf(n - i + 1)).divide(BigInteger.valueOf(i));
Listing 3–6 BigIntegerTest.java
1. import java.math.*;
2. import java.util.*;
3.
4. /**
5. * This program uses big numbers to compute the odds of winning the grand prize in a lottery.
6. * @version 1.20 2004-02-10
7. * @author Cay Horstmann
8. */
9. public class BigIntegerTest
10. {
11. public static void main(String[] args)
12. {
13. Scanner in = new Scanner(System.in);
14.
15. System.out.print("How many numbers do you need to draw? ");
16. int k = in.nextInt();
17.
18. System.out.print("What is the highest number you can draw? ");
19. int n = in.nextInt();
20.
21. /*
22. * compute binomial coefficient n*(n-1)*(n-2)*...*(n-k+1)/(1*2*3*...*k)
23. */
24.
25. BigInteger lotteryOdds = BigInteger.valueOf(1);
26.
27. for (int i = 1; i <= k; i++)
28. lotteryOdds = lotteryOdds.multiply(BigInteger.valueOf(n - i + 1)).divide(
29. BigInteger.valueOf(i));
30.
31. System.out.println("Your odds are 1 in " + lotteryOdds + ". Good luck!");
32. }
33. }

java.math.BigInteger

? BigInteger add(BigInteger other)
? BigInteger subtract(BigInteger other)
? BigInteger multiply(BigInteger other)
? BigInteger divide(BigInteger other)
? BigInteger mod(BigInteger other)
returns the sum, difference, product, quotient, and remainder of this big integer and
other.
? int compareTo(BigInteger other)
returns 0 if this big integer equals other, a negative result if this big integer is less
than other, and a positive result otherwise.
? static BigInteger valueOf(long x)
returns a big integer whose value equals x.
? BigDecimal add(BigDecimal other)
? BigDecimal subtract(BigDecimal other)
? BigDecimal multiply(BigDecimal other)
? BigDecimal divide(BigDecimal other, RoundingMode mode) 5.0
returns the sum, difference, product, or quotient of this big decimal and other.
To compute the quotient, you must supply a rounding mode. The mode
RoundingMode.HALF_UP is the rounding mode that you learned in school (i.e., round
down digits 0 . . . 4, round up digits 5 . . . 9). It is appropriate for routine
calculations. See the API documentation for other rounding modes.
? int compareTo(BigDecimal other)
returns 0 if this big decimal equals other, a negative result if this big decimal is less
than other, and a positive result otherwise.
? static BigDecimal valueOf(long x)
? static BigDecimal valueOf(long x, int scale)
returns a big decimal whose value equals x or x /10scale.

Brian 2010-08-17 13:21 发表评论

SGU 分类开��语

Brian — Tue, 17 Aug 2010 05:17:00 GMT

别看了，老子疯了。SGU的进展肯定相当缓慢，但是我会用心做的。下面先转一个大牛的Sgu袖珍分类�?br>101 Domino �Ƨ拉�?br>102 Coprime 枚�D/数学�Ҏ��
103 Traffic Lights 最短�\
104 Little Shop of Flowers 动态规�?br>105 Div 3 找规�?br>106 The Equation 扩展�Ƨ几里�d
107 987654321 Problem 找规�?br>108 Self-numbers II 枚�D+�{�法递推
109 Magic of David Copperfield II 构�?br>110 Dungeon 计算几何+模拟
111 Very Simple Problem 模拟�W�算开�?br>112 a^b-b^a 高精�?br>113 Nearly Prime Numbers 判质�?br>114 Telecasting Station 找中位数
115 Calendar 模拟
116 Index of Super-prime 动态规�?br>117 Counting 快速幂
118 Digital Root 模拟
119 Magic Pairs 枚�D
120 Archipelago 计算几何
121 Bridges Painting 构�?br>122 The Book 构造哈密顿回�\
123 The Sum 递推
124 Broken line 计算几何
125 Shtirlits 搜烦
126 Boxes 数学�Ҏ��
127 Telephone directory �l�计
128 Snake �q�查�?+ 树状数组
129 Inheritance 计算几何
130 Circle 卡特兰数
131 Hardwood floor 状态压�~�动�?br>132 Another Chocolate Maniac 状态压�~�动�?br>133 Border 贪心
134 Centroid 树型DP
135 Drawing Lines 找规�?br>136 Erasing Edges 数学�Ҏ��
137 Funny Strings 构�?br>138 Games of Chess 构�?br>139 Help Needed! 数学�Ҏ��
140 Integer Sequences 扩展�Ƨ几里�d
141 Jumping Joe 扩展�Ƨ几里�d
142 Keyword 枚�D
143 Long Live the Queen 树型DP
144 Meeting 数学�Ҏ��
145 Strange People 二分�{�案+搜烦
146 The Runner 数学�Ҏ��
147 Black-white King 枚�D
148 B-Station 枚�D+快排/二分/�?br>149 Computer Network 树型DP
150 Mr. Beetle II 枚�D
151 Construct a triangle 计算几何构�?br>152 Making round 构�?br>153 Playing With Matches 动态规�?扑��@环节
154 Factorial 二分数学�Ҏ��
155 Cartesian Tree 建笛卡尔�?br>156 Strange Graph �~�团+�Ƨ拉回�\
157 Patience 搜烦+开�?br>158 Commuter Train 枚�D
159 Self-replicating Numbers 扩展队列+高精�?br>160 Magic Multiplying Machine 动态规�?br>161 Intuitionistic Logic 搜烦*
162 Pyramids 数学�Ҏ��
163 Wise King 模拟
164 Airlines 贪心
165 Basketball 构�?br>166 Editor 模拟
167 I-country 动态规�?br>168 Matrix 动态规�?br>169 Numbers 数学�Ҏ��
170 Particles 贪心
171 Sarov zones 贪心
172 eXam 判断二分�?br>173 Coins 高斯消元
174 Wall �q�查�?Hash
175 Encoding 搜烦/动态规�?br>176 Flow construction 上下界网�l�流
177 Sqare 倒序染色
178 Chain 数学�Ҏ��
179 Brackets light 找规�?br>180 Inversions 归�ƈ排序/高��数据�l�构
181 X-Sequence 扑��@环节
182 Open the Brackets 搜烦
183 Painting the balls 动态规划优�?br>184 Patties 直接计算
185 Two shortest �|�络��?br>186 The chain 贪心
187 Twist and whirl -- want to cheat 块状链表
188 Factory guard 数学�Ҏ��
189 Perl-like Substr 模拟
190 Dominoes 二分囑֌��?br>191 Exhibition 贪心
192 RGB ��L��?�Q?�l�计
193 Chinese Girls' Amusement 数学�Ҏ�� + 高精�?br>194 Reactor Cooling �|�络��?br>195 New Year Bonus Grant 贪心
196 Matrix Multiplication 数学�Ҏ��
197 Nice Patterns Strike Back 动态规�?矩阵
198 Get Out! 计算几何
199 Beautiful People 最镉K��降子序列
200 Cracking RSA 高斯消元
201 Non Absorbing DFA 动态规�?br>202 The Towers of Hanoi Revisited 动态规划构�?br>203 Hyperhuffman 贪心
204 Little Jumper 二分+数学�Ҏ��*
205 Quantization Problem 动态规�?

Brian 2010-08-17 13:17 发表评论

国产精品s色,久久黄色影院,欧美主播一区二区三区

SGU 118 Digital Root

SGU 111 Very simple problem

SGU 107 987654321 problem

SGU 105 Div 3

SGU 102 Coprimes

SGU 113 Nearly prime numbers

SGU 112 a^b-b^a (Java Edition)

SGU 分类开���语

SGU 分类开��语