2014 ACM/ICPC Asia Regional Xi'an Online（hdu 5007 - hdu 5017）

1.Post Robot

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5007

解题思路：

题目大意：

给你一个文本，存在“Apple”, “iPhone”, “iPod”, “iPad”的字样输出”MAI MAI MAI!“，碰到”Sony“则输出“SONY DAFA IS GOOD!”

解题思路：

暴力模拟。。。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;char str[1000005];char tmp1[] = {"Apple"};char tmp2[] = {"iPhone"};char tmp3[] = {"iPod"};char tmp4[] = {"iPad"};char tmp5[] = {"Sony"};char ans1[] = {"MAI MAI MAI!"};char ans2[] = {"SONY DAFA IS GOOD!"};int main(){    int l;    while(gets(str)){        l = strlen(str);        for(int i = 0; i <= l-4; i++){            if(i <= l-5 && strncmp(str+i,tmp1,5) == 0)                printf("%s\n",ans1);            else if(i <= l-6 && strncmp(str+i,tmp2,6) == 0)                 printf("%s\n",ans1);            else if(strncmp(str+i,tmp3,4) == 0)                 printf("%s\n",ans1);            else if(strncmp(str+i,tmp4,4) == 0)                 printf("%s\n",ans1);            else if(strncmp(str+i,tmp5,4) == 0)                 printf("%s\n",ans2);        }    }    return 0;}

5.Game（博弈）

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5011

解题思路：

Nim博弈：http://blog.csdn.net/piaocoder/article/details/48472995

AC代码：

#include <iostream>#include <cstdio>using namespace std;int main(){    int n;    while(~scanf("%d",&n)){        int x,sum = 0;        for(int i = 0; i < n; i++){            scanf("%d",&x);            sum ^= x;        }        if(sum == 0)            printf("Lose\n");        else            printf("Win\n");    }    return 0;}

6.Dice（模拟+bfs）

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5012

解题思路：

题目大意：

给你两个色子，规定每个面的序号，这两个色子的每个面的颜色分别为：（a1,a2,a3,a4,a5,a5)(b1,b2,b3,b4,b5,b6)，现在要你通过

四种旋转使得a序列和b序列一样。可以的话输出最少的旋转步数，不能输出-1.

解题思路：

模拟即可。。。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;int aim;struct node{int s[10];int sum;int step;}a,b;int vis[700000];int base[10] = {0,100000,10000,1000,100,10,1};int bfs(){struct node n1, n2;int i,j,sum;queue<node>q;q.push(a);while(!q.empty()){n1 = q.front();q.pop();if(n1.sum == aim)  return n1.step;for(i = 1; i <= 4; i++){sum = 0;for(j = 1; j <= 6; j++)  n2.s[j] = n1.s[j];if(i == 1){n2.s[1] = n1.s[6];n2.s[2] = n1.s[5];n2.s[5] = n1.s[1];n2.s[6] = n1.s[2];}else if(i == 2){n2.s[1] = n1.s[5];n2.s[2] = n1.s[6];n2.s[5] = n1.s[2];n2.s[6] = n1.s[1];}else if(i == 3){n2.s[1] = n1.s[4];n2.s[2] = n1.s[3];n2.s[3] = n1.s[1];n2.s[4] = n1.s[2];}else{n2.s[1] = n1.s[3];n2.s[2] = n1.s[4];n2.s[3] = n1.s[2];n2.s[4] = n1.s[1];}for(j = 1; j <= 6; j++)  sum += n2.s[j] * base[j];if(!vis[sum]){n2.step = n1.step + 1;n2.sum = sum;vis[sum] = 1;q.push(n2);}}}return -1;}int main(){while(~scanf("%d%d%d%d%d%d",&a.s[1],&a.s[2],&a.s[3],&a.s[4],&a.s[5],&a.s[6])){memset(vis,0,sizeof(vis));int sum = 0;for(int i = 1; i <= 6; i++)  sum += a.s[i] * base[i];a.step = 0;a.sum = sum;vis[sum] = 1;for(int i = 1; i <= 6; i++)  scanf("%d",&b.s[i]);aim = 0;for(int i = 1; i <= 6; i++)  aim += b.s[i] * base[i];printf("%d\n",bfs());}return 0;}

8.Number Sequence（贪心+位处理）

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5014

解题思路：

贪心枚举即可。。。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <map>#include <algorithm>using namespace std;typedef long long ll;int a[100005];int b[100005];int vis[100005];int main(){    int n;    while(scanf("%d",&n) != EOF){        memset(vis,0,sizeof(vis));        for(int i = 0; i <= n; i++)            scanf("%d",&a[i]);        for(int i = n; i >= 0; i--){            int tmp = 0;            if(!vis[i]){                for(int j = 0; ; j++){                    if(!(i&(1<<j)))                        tmp += (1<<j);                    if(tmp >= i){                        tmp -= (1<<j);                        break;                    }                }                vis[i] = tmp;                vis[tmp] = i;            }        }        printf("%lld\n",(ll)(n)*(n+1));        for(int i = 0; i < n; i++)            printf("%d ",vis[a[i]]);        printf("%d\n",vis[a[n]]);    }    return 0;}

11.Ellipsoid

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5017

解题思路：

http://blog.csdn.net/piaocoder/article/details/39398507

AC代码：

#include <iostream>#include <cstdio>#include <cmath>using namespace std;const double INF = 0x3f3f3f3f;const double eps = 1e-9;const double r = 0.9;    //降温速度const int dx[] = {-1,0,1,-1,1,-1,0,1},dy[] = {1,1,1,0,0,-1,-1,-1};double a,b,c,d,e,f;double dis(double x,double y,double z){    return sqrt(x*x+y*y+z*z);}//已知x,y;求z(把x,y看做常量，z看做变量，求解普通的一元二次方程即可)double getZ(double x,double y){    double tmp = (d*y+e*x)*(d*y+e*x)-4*c*(a*x*x+b*y*y+f*x*y-1);    if(tmp < 0)        return 10;      //返回一个大于2的值即可，    double z1 = (-(d*y+e*x)+sqrt(tmp))/(2*c);    double z2 = (-(d*y+e*x)-sqrt(tmp))/(2*c);    if(z1*z1 <= z2*z2)        return z1;    else        return z2;}//模拟退火算法double solve(){    double step = 1;    double x = 0,y = 0,z,nx,ny,nz;    double ans = INF,sum;    int flag;    while(step > eps){        flag = 1;        while(flag){            flag = 0;            for(int i = 0; i < 8; i++){                nx = x+dx[i]*step;                ny = y+dy[i]*step;                nz = getZ(nx,ny);                if(nz > 2)        //此处应是一个大于1的值最好是2以上，因为double                    continue;                sum = dis(nx,ny,nz);                if(ans > sum){                    x = nx;y = ny;                    ans = sum;                    flag = 1;                }            }        }        step *= r;    }    return ans;}int main(){    while(scanf("%lf%lf%lf%lf%lf%lf",&a,&b,&c,&d,&e,&f)!=EOF)        printf("%.8lf\n",solve());}