UCF Local Programming Contest 2015

链接：https://www.jisuanke.com/contest/7330?view=challenges

A  Find the Twins

 

 

 题意：给一串数字看这对兄弟是否在其中。

需注意行末空格换行的格式问题

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 #define pi acos(-1)
 7 
 8 int a[15];
 9 
10 int main(){
11     int n;
12     scanf("%d", &n);
13     while(n --){
14         bool flag1 = false, flag2 = false;
15         for(int i = 0; i < 10; i ++ ){
16             scanf("%d", &a[i]);
17             if(a[i] == 18)
18                 flag1 = true;
19             else if(a[i] == 17)
20                 flag2 = true;
21             if(i != 9) printf("%d ", a[i]);
22             else printf("%d\n", a[i]);
23         }
24         if(flag1 && flag2) printf("both\n");
25         else if(flag1 && !flag2) printf("mack\n");
26         else if(!flag1 && flag2) printf("zack\n");
27         else printf("none\n");
28         if(n != 0) printf("\n");
29     }
30     return 0;
31 }
View Code

 

B  Medal Ranking

 

 

 

 

 

 

 

 题意：给出奖牌数，问哪一种计算奖牌排名的方法可以使USA赢

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 #define pi acos(-1)
 7 
 8 int a[10];
 9 
10 int main(){
11     int n;
12     scanf("%d", &n);
13     while(n --){
14         bool flag1 = false, flag2 = false;
15         int suma = 0, sumr = 0;
16         for(int i = 0; i < 6; i ++ ){
17             scanf("%d", &a[i]);
18             if(i != 5) printf("%d ", a[i]);
19             else printf("%d\n", a[i]);
20             if(i < 3)
21                 suma += a[i];
22             else
23                 sumr += a[i];
24         }
25         if(suma > sumr) flag1 = true;
26             if(a[0] > a[3]) flag2 = true;
27             else if(a[0] == a[3] && a[1] > a[4]) flag2 = true;
28             else if(a[0] == a[3] && a[1] == a[4] && a[2] > a[5]) flag2 = true;
29         
30         if(flag1 && flag2) printf("both\n");
31         else if(flag1 && !flag2) printf("count\n");
32         else if(!flag1 && flag2) printf("color\n");
33         else printf("none\n");
34         if(n != 0) printf("\n");
35     }
36     return 0;
37 }
View Code

 

C  Brownies vs. Candies vs. Cookies

 

 

 

 

 

 （题目好长

题意：老师可以不断将点心一分为二，需要保证在每队过来拿点心之后还有剩下的点心，不然队员就不训练了（

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 #define pi acos(-1)
 7 
 8 int a[10];
 9 
10 int main(){
11     int n;
12     scanf("%d", &n);
13     int flag = 0;
14     char s[10];
15     int sums, sumb, sumg, group;
16     while(n -- ){
17         flag ++ ;
18         scanf("%d %d", &sums, &sumb);
19         scanf("%d", &sumg);
20         printf("Practice #%d: %d %d\n", flag, sums, sumb);
21         while(sumg -- ){
22             scanf("%d", &group);
23             while(group >= sumb)
24                 sumb *= 2;
25             printf("%d %d\n", group, sumb - group);
26             sumb -= group;
27         }
28         printf("\n");
29     }
30     return 0;
31 }
View Code

 

D  Lemonade Stand

 

 

 

 

 

 题意：经营一个柠檬水摊点，提前知道以后每天会卖多少杯水、每天的柠檬价格和方糖的价格。求最少成本。

思路：记录每一天目前出现的最低价格，然后该天需要使用的柠檬和方糖在最低价格那天买即可。

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 #define pi 3.14159265358979
 7 
 8 int main(){
 9     int n;
10     scanf("%d", &n);
11     while(n --){
12         int nn, lemon, sugar;
13         int sum = 0;
14         int minl = 0x3f3f3f3f, mins = 0x3f3f3f3f;
15         double left = 0;
16         int num = 0;
17         scanf("%d %d %d", &nn, &lemon, &sugar);
18         for(int i = 0; i < nn; i ++ ){
19             int x, y;
20             int a, b, c;
21             int cnt = 0;
22             scanf("%d %d %d", &a, &b, &c);
23             minl = min(minl, b);
24             mins = min(mins, c);
25             x = a * lemon;
26             // while(num < a * sugar){
27             //     num += 80;
28             //     cnt++;
29             // }
30             // num = num - a * sugar;
31             double aa = (double)a * (double)sugar / 16.0 / 5.0 - left;
32             y = ceil(aa);
33             left = double(y) - aa;
34             sum += x * minl;
35             sum += y * mins;
36             //sum += cnt * mins;
37         }
38         printf("%d\n", sum);
39     }
40     return 0;
41 }
View Code

 

E  Rain Gauge

 

 

 

 

 

 一个简单的数学题~

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 #define pi 3.14159265358979
 7 
 8 int main(){
 9     int n;
10     scanf("%d", &n);
11     while(n --){
12         double d, r;
13         scanf("%lf %lf", &d, &r);
14         if(r >= d / sqrt(2)){
15             printf("%.2lf\n", d * d);
16         }
17         else if(r <= d / 2){
18             printf("%.2lf\n", pi * r * r);
19         }
20         else {
21             double x = acos(d / (2 * r));;
22             double cnt1 = (2 * pi - 8 * x) / 2 / pi;
23             double ans1 = cnt1 * pi * r * r;
24             double cnt2 = sqrt(r * r - (d/2) * (d/2));
25             double ans2 = cnt2 * d / 2 * 4;
26             printf("%.2lf\n", ans1 + ans2);
27         }
28     }
29     return 0;
30 }
View Code

 

posted @ 2020-03-14 21:06  moomight  阅读(309)  评论(0)  编辑  收藏  举报