Contest - 第10届“新秀杯”ACM程序设计大赛现场热身赛 赛后信息（题解）

	Problem Id	Title
	Problem A	A+B
	Problem B	统计字数
	Problem C	生日计算
	Problem D	冬瓜的寒假之旅

---

Problem A(略)

Problem B

B题目就是一个统计字符的简单模拟，纵向记录横向输出就可。先for一圈遍历遍最大的个数可确定高度。

 /****************************************/
 /*****            Desgard_Duan        *****/
 /****************************************/
 //#pragma comment(linker, "/STACK:102400000,102400000")
 #define _CRT_SECURE_NO_WARNINGS
 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <stack>
 #include <map>
 #include <queue>
 #include <vector>
 #include <set>
 #include <functional>
 #include <cmath>
 #include <numeric>

 using namespace std;

 vector<string> ans;
 int let[];

 int main () {
     string text = "", in;
     for (int i =  ; i < ; ++ i) {
         getline(cin, in);
         text += in;
     }
     //cout << text << endl;
     memset (let, , sizeof (let));
     int Max = ;
     for (int i = ; i < text.size(); ++ i) {
         if (isalpha(text[i])) {
             let[text[i] - 'A'] ++;
             Max = max (Max, let[text[i] - 'A']);
         }
     }
     for (int i = ; i <= Max; ++ i) {
         string text_line = "";
         for (int j =  ; j < ; ++ j) {
             if (i > Max - let[j]) {
                 text_line += "*";
             } else {
                 text_line += " ";
             }
             if (j != ) text_line += " ";
         }
         cout << text_line << endl;
     }
     puts("A B C D E F G H I J K L M N O P Q R S T U V W X Y Z");
     return ;
 }

Problem C

考虑闰年的2月29日生日就可，其他没有什么坑。

 /****************************************/
 /*****            Desgard_Duan        *****/
 /****************************************/
 //#pragma comment(linker, "/STACK:102400000,102400000")
 #define _CRT_SECURE_NO_WARNINGS
 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <stack>
 #include <map>
 #include <queue>
 #include <vector>
 #include <set>
 #include <functional>
 #include <cmath>
 #include <numeric>

 using namespace std;

 int judge( int x ) {
     if( x %  ==  || ( x %  ==  && x %   !=  ) )
         return ;
     return ;
 }
 int main() {
     int y, m, d, t;
     while(scanf( "%d", &t ) != EOF) {
         while( t-- ) {
             scanf( "%d-%d-%d", &y, &m, &d );
             int sum = ;
             if( judge( y ) && m ==  && d ==  && !judge( y +  ) ) {
                 puts( "-1" );
                 continue;
             }
             for( int i = ; i <= ; ++i )
                 if(judge( y + i ))
                     sum ++;
             if( judge( y +  ) && m <=  && d <  )
                 --sum;
             if( judge( y ) && m >=  )
                 --sum;
             printf( "%d\n", sum +  *  );
         }
     }
     return ;
 }

Problem D

主要考察枚举法。即全排列。在数据量不大的情况下，即可将TSP问题的所有情况全部列出，寻求最佳值。

另外可以学习一下next_permutation函数的用法，白书中也有介绍。

 #include <cstring>
 #include <iostream>
 #include <cstdlib>
 #include <cctype>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <map>
 #include <set>
 #include <queue>

 using namespace std;

 int per[];

 struct point {
     double x, y;
 } p[];

 double dis(point a, point b) {
     return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
 }

 int main() {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T, n, cas = ;
     double ans;
     bool f;
     scanf("%d", &T);
     while(T--) {
         f = true;
         scanf("%d", &n);
         for(int i = ; i <= n; ++i) {
             scanf("%lf%lf", &p[i].x, &p[i].y);
         }
         for(int i = ; i <= n; ++i) {
             per[i - ] = i;
         }
         do {
             point tmp;
             tmp.x = ;
             tmp.y = ;
             double sum = dis(tmp, p[per[]]);
             for(int i = ; i < n; ++i) {
                 sum += dis(p[per[i]], p[per[i - ]]);
             }
             sum += dis(tmp, p[per[n - ]]);
             if(f) {
                 ans = sum;
                 f = false;
             } else
                 ans = min(ans, sum);
         } while(next_permutation(per, per + n));
         printf("Case #%d: %.2lf\n", cas++, ans);
     }
     return ;
 }

---