04.06 UCF Local Programming Contest 2017

A.Electric Bill

题意：简单计算题，超过1000部分额外算

 1 #include<stdio.h>
 2 int main(){
 3     int money1,money2;
 4     long long int sum=0,n,num;
 5     scanf("%d %d",&money1,&money2);
 6     scanf("%lld",&n);
 7     while(n--){
 8             sum=0;
 9         scanf("%lld",&num);
10         if(num>1000){
11             sum+=(num-1000)*money2;
12             sum+=1000*money1;
13             printf("%lld %lld\n",num,sum);
14         }else{
15             printf("%lld %lld\n",num,num*money1);
16         }
17     }
18 }

B.Simplified Keyboard

题目：判断是题目中说的哪一类

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int pos[9] = {-10,-9,-8,-1,0,1,8,9,10};
 4 char a[27] = "abcdefghijklmnopqrstuvwxyz";
 5 int main()
 6 {
 7     int n;
 8     scanf("%d",&n);
 9     while(n--)
10     {
11         int i,j,k;
12         int flag = 3;
13         char s1[21],s2[21];
14         scanf("%s %s",s1,s2);
15         if(strcmp(s1,s2)==0)
16             flag=1;
17         else
18         {
19             if(strlen(s1)!=strlen(s2))
20                 flag = 3;
21             else
22             {
23
24                 flag = 2;
25                 for(i = 0; i < strlen(s1); i++)
26                 {
27                     int isok = 0;
28                     for(j = 0; j < 9; j++)
29                     {
30                         if(s2[i]-'a'+pos[j]>=0&&s2[i]-'a'+pos[j]<=25)
31                         {
32                             if(s1[i]==a[s2[i]-'a'+pos[j]])
33                             {
34                                 isok = 1;
35                                 break;
36                             }
37
38
39                         }
40
41                     }if(isok == 1)
42                             continue;
43                             else {flag = 3;break;}
44
45                 }
46
47             }
48         }
49         printf("%d\n",flag);
50     }
51
52
53
54 }

C.Singin' in the Rain

题目：和上场比赛的CDs题有点相同，只是贪心找最小即可，把样例一列出来，找找规律即可

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<algorithm>
 4 #include<iostream>
 5 using namespace std;
 6 int main(){
 7     int m,n,t,s;
 8     long long int sum=0;
 9     int a[1010];
10     scanf("%d",&n);
11     while(n--){
12         sum=0;
13         scanf("%d %d",&t,&s);
14         for(int i=0;i<s;i++){
15             scanf("%d",&a[i]);
16         }
17         for(int i=0;i<s-1;i++){
18             if(a[i]>a[i+1]){
19                 sum+=min((a[i]-a[i+1]+1),(a[i+1]+t-1-a[i]));
20             }else if(a[i]==a[i+1]){
21                 sum++;
22             }else if(a[i]+1==a[i+1]){
23                 continue;
24             }else{
25                 sum+=min((a[i+1]-a[i]-1),(a[i]+t-a[i+1]+1));
26             }
27         }
28         printf("%lld\n",sum);
29     }
30 }

E.Simple Darts

题目：进行区域判断，通过反三角函数和圆位置进行相对判断，从而确定分数

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<string>
 6 #include<map>
 7 #include<sstream>
 8 #include<cstring>
 9 #include<vector>
10 #include<queue>
11 #define LL long long
12 const double pi=3.1415926;
13 using namespace std;
14 int main(){
15     int n;
16     scanf("%d",&n);
17     while(n--){
18         int w,b,d,s;
19         cin >> w >> b >> d >> s;
20         int t;
21         cin >> t;
22         long long int ans = 0;
23         while(t--){
24             double x,y;
25             cin >> x >> y;
26             double r,jiao;
27             r = x*x + y*y;
28             double fen = 2*pi/w;
29             if(r < b*b){
30                 ans+=50;
31             }else if(r > b*b && r < d*d){
32                 if((y/x)>0&&y>0){
33                     jiao=atan(y/x);
34                 }else if((y/x)>0&&y<0){
35                     jiao=atan(y/x)+pi;
36                 }else if((y/x)<0&&y<0){
37                     jiao=atan(y/x)+2*pi;
38                 }else if((y/x)<0&&x<0){
39                     jiao=atan(y/x)+pi;
40                 }else if(x==0&&y>0){
41                     jiao=pi/2;
42                 }else if(x==0&&y<0){
43                     jiao=(pi*3)/2;
44                 }else if(y==0&&x<0){
45                     jiao=pi;
46                 }else if(y==0&&x>0){
47                     jiao=0;
48                 }
49                 int j=jiao/fen;
50                 ans+=((j+1)*2);
51             }else if(r > d*d && r < s*s){
52               if((y/x)>0&&y>0){
53                     jiao=atan(y/x);
54                 }else if((y/x)>0&&y<0){
55                     jiao=atan(y/x)+pi;
56                 }else if((y/x)<0&&y<0){
57                     jiao=atan(y/x)+2*pi;
58                 }else if((y/x)<0&&x<0){
59                     jiao=atan(y/x)+pi;
60                 }else if(x==0&&y>0){
61                     jiao=pi/2;
62                 }else if(x==0&&y<0){
63                     jiao=(pi*3)/2;
64                 }else if(y==0&&x<0){
65                     jiao=pi;
66                 }else if(y==0&&x>0){
67                     jiao=0;
68                 }
69                 int j=jiao/fen;
70                 ans+=(j+1);
71             }else{
72                 ans += 0;
73             }
74         }
75         printf("%lld\n",ans);
76     }
77 }