MCPC 2011Hdu4207-4214（未完全）题解

第一题：Grade School Multiplication

http://acm.hdu.edu.cn/showproblem.php?pid=4207

题目大意：给你两个数，要求按照题目所讲的规律一步步得出结果。

如：给你两个数432 和 5678,你就要按照下面的规律输出

432

5678

-------

3456

3024

2592

2160

-------

2452896

模拟。

因为数据范围不是很大，所以直接用long long型计算每一位然后输出结果就行。

注意：特别要注意0的情况，因为没有考虑这种情况导致了多次wa。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <queue>
using namespace std;
template <class T> void checkmin(T &t,T x) {if(x < t) t = x;}
template <class T> void checkmax(T &t,T x) {if(x > t) t = x;}
template <class T> void _checkmin(T &t,T x) {if(t==-1) t = x; if(x < t) t = x;}
template <class T> void _checkmax(T &t,T x) {if(t==-1) t = x; if(x > t) t = x;}
typedef pair <int,int> PII;
typedef pair <double,double> PDD;
typedef long long ll;
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end ; it ++)
ll a , b;
int cas = 1;
int wei(ll a) {
    int ans = 0;
    while(a) {
        a /= 10;
        ans ++;
    }
    checkmax(ans , 1);
    return ans;
}
int oowei(ll a) {
    int ans = 0;
    while(a) {
        if(a % 10 != 0) ans ++;
        a /= 10;
    }
    checkmax(ans , 1);
    return ans;
}
int main() {
    while(cin >> a >> b) {
        if(a + b == 0) break;
        printf("Problem %d\n",cas++);
        ll c = a * b;
        int Wa = wei(a) , Wb = wei(b) , Wc = wei(c);
        if(c == 0) {
            int dd = max(Wa , Wb);
            if(a != 0) cout << a << endl;
            else for(int i=1;i<dd;i++) putchar(' ');
            cout << a << endl;
            if(b != 0) cout << b << endl;
            else for(int i=1;i<dd;i++) putchar(' ');
            cout << b << endl;
            for(int i=0;i<dd;i++) putchar('-');
            puts("");
            for(int i=1;i<dd;i++) putchar(' ');
            cout << c << endl;
            continue;
        }
        int oWb = oowei(b);
        for(int i=0;i<Wc-Wa;i++) putchar(' ');
        cout << a << endl;
        for(int i=0;i<Wc-Wb;i++) putchar(' ');
        cout << b << endl;
        for(int i=0;i<Wc;i++) putchar('-');
        puts("");
        int cc = 0;
        ll ad = 0;
        while(b) {
            ll d = b%10;
            b /= 10;
            if(d == 0) { ad +=1; cc++; continue; }
            d *= a;
            ll Wd = wei(d);
            for(int i=0;i<Wc-Wd-cc;i++) putchar(' ');
            cout << d ;
            while(ad > 0) putchar('0') , ad -= 1;
            cout << endl;
            cc ++;
        }
        if(oWb > 1) {
            for(int i=0;i<Wc;i++) putchar('-');
            puts("");
            cout << c << endl;
        }
    }
    return 0;
}

第二题：Laser Tag

http://acm.hdu.edu.cn/showproblem.php?pid=4208

题目大意：有n面镜子（ 1 ≤ n ≤ 7），现在我站在(0,0)点发射激光，激光照到镜子上会按照物理规律反射，最后有可能射到我自己；求沿x轴正方向0->359度所有能伤到自己的发射角度。

计算几何。

（题解：暂缺）

第三题：Pizza Pricing

http://acm.hdu.edu.cn/showproblem.php?pid=4209

题目大意：简单题。找价值最高的pizza，输出他的直径。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <queue>
using namespace std;
template <class T> void checkmin(T &t,T x) {if(x < t) t = x;}
template <class T> void checkmax(T &t,T x) {if(x > t) t = x;}
template <class T> void _checkmin(T &t,T x) {if(t==-1) t = x; if(x < t) t = x;}
template <class T> void _checkmax(T &t,T x) {if(t==-1) t = x; if(x > t) t = x;}
typedef pair <int,int> PII;
typedef pair <double,double> PDD;
typedef long long ll;
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end ; it ++)
int n , a , b , ans , cas = 1;
double tmp;
int main() {
    while(~scanf("%d",&n) && n) {
        tmp = -1.0;
        for(int i=0;i<n;i++) {
            scanf("%d%d",&a,&b);
            double tp = ((double)(a*a)) / ((double)b);
            if(tp > tmp) {ans = a; tmp = tp;}
        }
        printf("Menu %d: %d\n",cas++,ans);
    }
    return 0;
}

第四题：Su-domino-ku

http://acm.hdu.edu.cn/showproblem.php?pid=4210

题目大意：数独题目的变形。要求拼出符合如下规律的数独：
Each row must contain each of the digits 1 through 9.
Each column must contain each of the digits 1 through 9.
Each of the indicated three-by-three squares must contain each of the digits 1 through 9.
For a su-domino-ku, nine arbitrary cells are initialized with the numbers 1 to 9. This leaves 72 remaining cells. Those must be filled by making use of the following set of 36 domino tiles. The tile set includes one domino for each possible pair of unique numbers from 1 to 9 (e.g., 1+2, 1+3, 1+4, 1+5, 1+6, 1+7, 1+8, 1+9, 2+3, 2+4, 2+5, ...). Note well that there are not separate 1+2 and 2+1 tiles in the set; the single such domino can be rotated to provide either orientation. Also, note that dominos may cross the boundary of the three-by-three squares (as does the 2+9 domino in our coming example).

DLX解决精确覆盖问题。

代码：

第五题：Refrigerator Magnets

http://acm.hdu.edu.cn/showproblem.php?pid=4211

题目大意：输入多串数，如果某个串中没有那个字母出现超过一次，则输出。

简单题。直接check一下就行。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <queue>
using namespace std;
template <class T> void checkmin(T &t,T x) {if(x < t) t = x;}
template <class T> void checkmax(T &t,T x) {if(x > t) t = x;}
template <class T> void _checkmin(T &t,T x) {if(t==-1) t = x; if(x < t) t = x;}
template <class T> void _checkmax(T &t,T x) {if(t==-1) t = x; if(x > t) t = x;}
typedef pair <int,int> PII;
typedef pair <double,double> PDD;
typedef long long ll;
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end ; it ++)
int main() {
    char ch[100100];
    bool vis[26];
    while(gets(ch)) {
        if(strcmp(ch,"END") == 0) break;
        int ok = 1;
        memset(vis,0,sizeof(vis));
        for(int i=0;ch[i];i++) {
            if(ch[i] == ' ') continue;
            int t = ch[i] - 'A';
            if(vis[t]) { ok=0;break; }
            vis[t] = 1;
        }
        if(ok) puts(ch);
    }
    return 0;
}