CF 239

C 

http://codeforces.com/contest/408/problem/C

题意：给一个直角三角行的两条直角边a,b，求出直角三角形的三个顶点（必须是整数）满足该直角三角形的三条边都不平行于坐标轴。

思路：对于a,b,若它不平行于坐标轴，那么它肯定是某两个整数的平方和。然后根据点积判断这两条假设的直角边是否垂直，垂直的前提下判断第三条边是否平行于坐标轴。若a,b不是某两个整数的平方和输出NO、

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;

const int INF = 0x3f3f3f3f;

int p[1010];
struct node
{
    int x,y;
}aa[1010],bb[1010];

void init()
{
    for(int i = 1; i <= 1000; i++)
    {
        p[i] = i*i;
    }
}

int solve(int x1,int y1,int x2,int y2)
{
    if(x1 * x2 + y1 * y2  == 0)
        return 1;
    return 0;
}

int main()
{
    init();
    int a,b;


    while(~scanf("%d %d",&a,&b))
    {
        int t = max(a,b);
        int cnt1 = 0;
        int cnt2 = 0;

        for(int i = 1; i <= t; i++)
        {
            for(int j = 1; j <= t; j++)
            {
                if(p[i] + p[j] == a*a)
                {
                    aa[cnt1++] = (struct node){i,j};
                }
                if(p[i] + p[j] == b*b)
                {
                    bb[cnt2++] = (struct node){i,j};
                }
            }
        }


        if(cnt1 == 0 || cnt2 == 0)
        {
            printf("NO\n");
            continue;
        }

        int f = 0;

        for(int i = 0; i < cnt1; i++)
        {
        	int x1 = aa[i].x;
        	int y1 = aa[i].y;
            for(int j = 0; j < cnt2; j++)
            {
            	int x2 = bb[j].x;
            	int y2 = bb[j].y;

                if(solve(-x1,y1,x2,y2) == 1)
                {
                    if(y1 != y2)
                    {
                        printf("YES\n");
                        printf("0 0\n");
                        printf("%d %d\n",-x1,y1);
                        printf("%d %d\n",x2,y2);
						f = 1;
						break;
                    }
                }
                if(solve(-x1,y1,y2,x2) == 1)
                {
                    if(y1 != x2)
                    {
                        printf("YES\n");
                        printf("0 0\n");
                        printf("%d %d\n",-x1,y1);
                        printf("%d %d\n",y2,x2);
                        f = 1;
						break;
                    }
                }
                if(solve(-y1,x1,x2,y2) == 1)
                {
					if(x1 != y2)
                    {
                        printf("YES\n");
                        printf("0 0\n");
                        printf("%d %d\n",-y1,x1);
                        printf("%d %d\n",x2,y2);
                        f = 1;
						break;
                    }
                }
                if(solve(-y1,x1,y2,x2) == 1)
                {
                    if(x1 != x2)
                    {
                        printf("YES\n");
                        printf("0 0\n");
                        printf("%d %d %d %d\n",-y1,x1,y2,x2);
						f = 1;
						break;
                    }
                }
            }
            if(f)
				break;
        }
		if(f == 0) printf("NO\n");
    }
    return 0;

}

D. Long Path 

http://codeforces.com/problemset/problem/407/B

题意：有n+1个迷宫，编号1~n+1，Vasya从1开始，初始他在1号房间标记1，他按如下规则前进：

他每走进一个房间，都会在房间上多画一个十字架；

假设他当前在i房间，已画上十字架，如果天花板上有奇数个十字架，他要移动到pi房间，否则就移动到i+1房间。

问他走到n+1房间要走的步数。。。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define LL __int64
using namespace std;

const int mod = 1000000007;
int n;
LL dp[1010];
int a[1010];

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i = 1; i <= n; i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        LL ans = 0;

        dp[1] = 2; 
        ans += dp[1];

        for(int i = 2; i <= n; i++)
        {
            dp[i] = 2;
            for(int j = a[i]; j < i; j++)
                dp[i] = (dp[i]+dp[j])%mod;

            ans = (ans%mod + dp[i])%mod;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}