Codeforces Round #278 (Div. 2)

A

给出一个数a, b为一个正数，如果a+b含有8,则称为吉祥的数字，找出最小的b

枚举b就可以了-

 1 #include<iostream>  
 2 #include<cstdio>  
 3 #include<cstring>  
 4 #include<algorithm>  
 5 using namespace std;
 6 
 7 int num[10]={1000000000,100000000,10000000,1000000,100000,10000,1000,100,10};
 8 
 9 int judge(long long x)
10 {
11     if(x<0)  x=-x;
12     
13     if(x/num[0]==8||x/num[1]%10==8||x/num[2]%10==8||x/num[3]%10==8||x/num[4]%10==8||x/num[5]%10==8||x/num[6]%10==8||x/num[7]%10==8||x/num[8]%10==8||x%10==8)
14     return 1;
15     return 0;
16 }
17 int main()
18 {
19     long long a,x,y,i;
20     int flag=1;
21     cin>>a;
22     for(i=1;i<=1000000000;i++)
23     {
24         x=a+i;
25         if(judge(x)||judge(y))
26         {
27             flag=1;
28             break;
29         }
30         
31     }
32     if(flag) printf("%I64d\n",i);
33     else printf("-1\n");
34 }
View Code

 B

现在给出n（0<=n<=4），a1,a2,--,an, 判断该盒子是否满足上述关系，如果满足，输出其余盒子的值

由上面的关系可以算出 a1+a4=a2+a3; a4=3*a1;

又因为题目里面说了给出的盒子的值不一定是他们的顺序，所以枚举可能地顺序，依次判断是否满足关系

  1 #include<iostream>  
  2 #include<cstdio>  
  3 #include<cstring>  
  4 #include<algorithm>  
  5 using namespace std;
  6 
  7 int a[15],b[15];
  8 int judge()
  9 {
 10     int x=a[1]+a[4]-a[2]-a[3];
 11     int y=a[4]/a[1];
 12     if(x==0&&y==3) return 1;
 13     return 0;
 14 }
 15 int check()
 16 {
 17     if(a[4]-a[3]>=0&&a[3]-a[2]>=0&&a[2]-a[1]>=0) return 1;
 18     return 0;
 19 }
 20 
 21 int main()
 22 {
 23     int i,j,n,x,y,z,flag=1;
 24     scanf("%d",&n);
 25     if(n==0)
 26     {
 27      printf("YES\n");
 28      printf("1\n1\n3\n3\n");    
 29     }
 30     if(n==1)
 31     {
 32         scanf("%d",&x);
 33         printf("YES\n");
 34         printf("%d\n%d\n%d\n",x,3*x,3*x);
 35         
 36     }
 37     if(n==2)
 38     {
 39         scanf("%d %d",&x,&y);int c=min(x,y);int d=max(x,y);
 40         x=c;y=d;
 41         for(i=1;i<=6;i++)
 42         {
 43             if(i==1)
 44             {
 45                 a[1]=x;a[2]=y;a[3]=4*x-y;a[4]=3*x;
 46             //    printf("i=%d\n",i);
 47                 if(judge()&&check()) {printf("YES\n");printf("%d\n%d\n",a[3],a[4]);flag=1;;break;}
 48                 else {flag=0;}
 49             }
 50             if(i==2)
 51             {
 52                 a[1]=x;a[3]=y;a[2]=4*x-y;a[4]=3*x;
 53             //    printf("i=%d\n",i);
 54                 if(judge()&&check()) {printf("YES\n");printf("%d\n%d\n",a[2],a[4]);flag=1;;break;}
 55                 else flag=0;
 56             }
 57             if(i==3)
 58             {
 59                 a[1]=x;a[4]=y;
 60             //    printf("i=%d\n",i);
 61                 for(j=x;j<=y;j++)
 62                 {
 63                     a[2]=j;a[3]=x+y-j;
 64                     if(judge()&&check()) {printf("YES\n");printf("%d\n%d\n",a[2],a[3]);flag=1;break;}
 65                     else flag=0;
 66                 }                
 67             }
 68             if(i==4)
 69             {
 70                 a[2]=x;a[3]=y;
 71             //    printf("i=%d\n",i);
 72                 for(j=1;j<=x;j++)
 73                 {
 74                     a[1]=j;a[4]=3*a[1];
 75                     if(judge()&&check()) {printf("YES\n");printf("%d\n%d\n",a[1],a[4]);flag=1;break;}
 76                     else flag=0;
 77                     
 78                 }
 79             }
 80             if(i==5)
 81             {
 82                 a[2]=x;a[4]=y;a[1]=y/3;a[3]=a[1]+a[4]-a[2];
 83             //    printf("-----------");
 84             //    printf("i=%d\n",i);
 85                 if(judge()&&check()) {printf("YES\n");printf("%d\n%d\n",a[1],a[3]);flag=1;break;}
 86                 else flag=0;
 87             }
 88             if(i==6)
 89             {
 90                 a[3]=x;a[4]=y;a[1]=y/3;a[2]=a[1]+a[4]-a[3];
 91             //    printf("i=%d\n",i);
 92             //    printf("flag=%d\n",flag);
 93                 if(judge()&&check()) {printf("YES\n");printf("%d %d\n",a[1],a[2]);flag=1;break;}
 94                 else flag=0;
 95             }                        
 96         }
 97         if(flag==0) {
 98                 printf("NO\n");
 99             }        
100     }
101     if(n==3)
102     {
103         scanf("%d %d %d",&b[1],&b[2],&b[3]);sort(b+1,b+4);
104         for(i=1;i<=4;i++)
105         {
106             if(i==1)
107             {
108                 a[1]=b[1];a[2]=b[2];a[3]=b[3];a[4]=3*a[1];
109             //    printf("i=%d\n",i);
110                 if(judge()&&check()) {printf("YES\n");printf("%d\n",a[4]);flag=1;break;}
111                 else flag=0;
112             }
113             if(i==2)
114             {
115                 a[1]=b[1];a[2]=b[2];a[4]=b[3];a[3]=a[1]+a[4]-a[2];
116             //    printf("i=%d\n",i);
117                 if(judge()&&check()) {printf("YES\n");printf("%d\n",a[3]);flag=1;break;}
118                 else flag=0;            
119             }
120             if(i==3)
121             {
122                 a[1]=b[1];a[3]=b[2];a[4]=b[3];a[2]=a[1]+a[4]-a[3];
123                 if(judge()&&check()) {printf("YES\n");printf("%d\n",a[2]);flag=1;break;}
124                 else flag=0;                
125             }
126             if(i==4)
127             {
128                 a[2]=b[1];a[3]=b[2];a[4]=b[3];a[1]=b[3]/3;
129                 if(judge()&&check()) {printf("YES\n");printf("%d\n",a[1]);flag=1;break;}
130                 else flag=0;
131             }
132             
133         }
134             if(flag==0) {
135                 printf("NO\n");
136             }    
137     }
138     if(n==4)
139     {
140         scanf("%d %d %d %d",&a[1],&a[2],&a[3],&a[4]);
141         sort(a+1,a+5);
142         if(judge()&&check()) printf("YES\n");
143         else printf("NO\n");
144     }
145     return 0;
146 }
View Code

 

 

 

话说这一题= =是既写得挫= =还改得久= =啊 啊 啊

posted @ 2015-02-27 11:19  sequenceaa  阅读(127)  评论(0)  编辑  收藏  举报