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nk_ysg — Sat, 02 Jun 2012 06:31:00 GMT

/*
*/
#include
using namespace std;

/*
注意到对于gcd(a,b) = d 我们�?a, b)用欧几里徯��转相除会最�l�得�?br />(d, 0)此时对于把a =d, b = 0 带入a*x + b*y = d�Q�显然x = 1�Q�y可以��Z�Q意��|��
�q�里y可以��Z�Q意值就意味着解会有无��C��。我们可以用a = d, b = 0的情况逆推出来
��M��gcd(a, b) = d 满��a*x + b*y = d的解。如果x0, y0是b*x + (a%b)*y = d 的解�Q?br />那么对于a*x + b*y = d的解呢？
b*x0 + (a%b)*y0 = d => b*x0 + (a - [a/b]*b)*y0 = a*y0 + b*(x0 - [a/b]*y0)�Q?br />所以a*x + b*y = d的解x1 = y0�Q?nbsp;y1 = x0 - [a/b]*y0; �q�样我们可以�E�序�q�带了�?br />*/
int extEuclid(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int d = extEuclid(b, a % b, x, y);
    int iTemp = x;
    x = y;
    y = iTemp - (a / b)* y;
    return d;
}
//解同余方�E�ax = b(mod n) (�q�回最��的正数x)

int modularLinearEquation(int a, int b, int n)
{
    //�{��h于求ax + cn = b;
    //先求a*x1 + c1*n = gcd(a, n)
    int x, y, d;
    d = extEuclid(a, n, x, y);
    if (b % d != 0)
       return -1;
    x = x * (b / d);
    x = (( x % n) + n) % n;
    return x;
}

//中国剩余定理�Q�推导都是数�?/span>
int solModularEquations(int b[], int m[], int k)
{

    int iTemp;
    int y;
    int result;

    int M = 1;
    for (int i = 0; i < k; i++)
      M *= m[i];

    result = 0;
    for (int i = 0; i < k; i++)
    {
      iTemp = M / m[i];
      y = modularLinearEquation(iTemp, 1, m[i]);
      result = (result + b[i] * iTemp * y) % M;
    }
    return result;
}

int main()
{
    int x, y , d;
    d = extEuclid(1001, 767, x, y);
    cout << x << endl;
    cout << y << endl;
    cout << d << endl;
    cout << "1001 * x + 767 * y = " << (1001 * x + 767 * y) << endl;
    cout << modularLinearEquation(3, 2, 100) << endl;
    return 0;
}

nk_ysg 2012-06-02 14:31 发表评论

数字三角�?记忆化搜索和递归)

nk_ysg — Fri, 07 Oct 2011 17:24:00 GMT

/*
�l�定一个由n行数字组成的数字三角形如下图所�C��?br />试设计一个算法，计算��Z��三角形的��至底的一条�\径，使该路径�l�过的数字��d��
对于�l�定的由n行数字组成的数字三角形，
�~�程计算从三角�Ş的顶臛_��的�\径经�q�的数字和的最大�?br />
*/
#include <stdio.h>
#include <string.h>

const int MAXN = 100 + 10;

int a[MAXN][MAXN],d[MAXN][MAXN];
int n;

inline int max(int a,int b) { return a > b ? a : b;}

int dfs(int i,int j)
{
    if (d[i][j] >= 0) return d[i][j];
    return d[i][j] = a[i][j] + (i == n ? 0 : max(dfs(i+1,j),dfs(i+1,j+1)));
}

int main()
{
    int i;
    int j;
    freopen("1.txt","r",stdin);
    while (scanf("%d",&n) != EOF)
    {
        memset(d,-1,sizeof(d));
        for (i = 1; i <= n; ++i)
        {
            for (j = 1; j <= i; ++j)
                scanf("%d",&a[i][j]);
        }
        printf("%d\n",dfs(1,1));
    }
    return 0;
}

#include <stdio.h>

const int MAXN = 100 + 10;

int a[MAXN][MAXN],d[MAXN][MAXN];

int n;

inline int max(int a, int b) { return a > b ? a : b; }

int main()
{
    int i,j;
    freopen("1.txt","r",stdin);
    while (scanf("%d",&n) != EOF)
    {
        for (i = 1; i <= n; ++i)
            for (j = 1; j <= i; ++j)
                scanf("%d",&a[i][j]);

        for (j = 1; j <= n; j++) d[n][j] = a[n][j];
        for (i = n - 1; i >= 1; i--)
            for (j = 1; j <= i; j++)
                d[i][j] = a[i][j] + max(d[i+1][j],d[i+1][j+1]);
        printf("%d\n",d[1][1]);
    }

    return 0;
}

nk_ysg 2011-10-08 01:24 发表评论

分数拆分

nk_ysg — Fri, 07 Oct 2011 09:04:00 GMT

1 /*
2 输入正整数k,扑ֈ�所有的正整数x>=y,使得1/k = 1/x + 1/y.
3 样例输入:
4 2
5 12
6 样例输出
7 2
8 1/2=1/6+1/3
9 1/2=1/4+1/4
10
11 8
12 1/12 = 1/156 + 1/13
13 1/12 = 1/84 + 1/14
14 1/12 = 1/60 + 1/15
15 1/12 = 1/48 + 1/16
16 1/12 = 1/36 + 1/18
17 1/12 = 1/30 + 1/20
18 1/12 = 1/28 + 1/21
19 1/12 = 1/24 + 1/24
20 */
21 #include <stdio.h>
22
23 const int MAXN = 1000 + 10;
24 int x[MAXN],y[MAXN];
25
26 int main()
27 {
28     int k;
29     int ans;
30     int i;
31     while (scanf("%d",&k) != EOF)
32     {
33         ans = 0;
34         for (i = k + 1; i <= 2*k; ++i)
35         {
36             if ((k * i) % (i-k) == 0) { x[ans] = (k*i)/(i-k),y[ans++] = i; }
37         }
38         printf("%d\n",ans);
39         for (i = 0; i < ans; ++i)
40         {
41             printf("1/%d = 1/%d + 1/%d\n",k,x[i],y[i]);
42         }
43     }
44     return 0;
45 }
46

nk_ysg 2011-10-07 17:04 发表评论

拓扑排序(感谢RoBa和NOCOW)

nk_ysg — Thu, 06 Oct 2011 11:19:00 GMT

Genealogical tree. 家谱�? Time Limit: 1 second Memory Limit: 16M

Background 大意�Q�火星�h的种族关�p�L��较�؜乱。宗族管理委员会�q�行开会发�a��{�活动时�Q�又想遵守先长辈�Q�后晚辈的顺序�?于是有下面的问题�Q?

Problem �~�写一个程序对�l�定的�h员，军_��一个先后顺序， �q�个��序必须遵守先长辈，后晚辈的原则�?

Input 首行为N�Q?1 <= N <= 100 ,N为��M�h数。根据百�q�传�l�，�l��h员用自然数编号�ؓ1至N。以下的N行，�W�I行�ؓ�W�I个�h的子孙列表，子孙的顺序是��L��的，用空格隔开�Q�且�?为结束。子孙列表可以是�I�的�?

Output 在一行内输出发言的顺序�?如有多个��序满��条�g�Q�则输出��M��个。至��会有一个满��的��序的�?

Sample Input

5 0 4 5 1 0 1 0 5 3 0 3 0

Sample Output

2 4 5 3 1

/* URAL 1022 Genealogical tree
* 900803 09:04:18 21 Aug 2005 RoBa 1022 C++ Accepted 0.001 190 KB
* 拓扑排序
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

const int MAX = 101;
int map[MAX][MAX],inde[MAX],taken[MAX];

int main()
{
    int i,n,tmp,j;
    while (scanf("%d",&n)!=EOF)
    {
        memset(map,0,sizeof(map));
        memset(inde,0,sizeof(inde));
        for (i = 1 ; i <= n ; i++)
            while (scanf("%d",&tmp),tmp)
            {
                map[i][tmp] = 1;
                ++inde[tmp];
            }
        while (1)
        {
            for (i = 1 ; i <= n ; i++)
                if (inde[i] == 0 && taken[i] == 0) break;
            if (i > n) break;
            for (j = 1 ; j <= n ; j++)
                if (map[i][j])
                {
                    map[i][j] = 0;
                    --inde[j];
                }
            taken[i] = 1;
            printf("%d ",i);
        }
        printf("\n");
    }
    return 0;
}

nk_ysg 2011-10-06 19:19 发表评论