2017ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）HDU6225.Little Boxes-大数加法

Posted 2020-11-10 Persistent.

整理代码。。。

Little Boxes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2304    Accepted Submission(s): 818

Problem Description

Little boxes on the hillside.
Little boxes made of ticky-tacky.
Little boxes.
Little boxes.
Little boxes all the same.
There are a green boxes, and b pink boxes.
And c blue boxes and d yellow boxes.
And they are all made out of ticky-tacky.
And they all look just the same.

 

 

Input

The input has several test cases. The first line contains the integer t (1 ≤ t ≤ 10) which is the total number of test cases.
For each test case, a line contains four non-negative integers a, b, c and d where a, b, c, d ≤ 2^62, indicating the numbers of green boxes, pink boxes, blue boxes and yellow boxes.

 

 

Output

For each test case, output a line with the total number of boxes.

 

 

Sample Input

4

1 2 3 4

0 0 0 0

1 0 0 0

111 222 333 404

 

 

Sample Output

10

0

1

1070

 

 

Source

2017ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

 

 

水题

代码:

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<cstdlib>
 7 #include<queue>
 8 #include<stack>
 9 #include<vector>
10 using namespace std;
11 typedef long long ll;
12 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
13 const int maxn=25;
14 const int mod=1e9+7;
15 const int inf=0x3f3f3f3f;
16 const double eps=acos(-1.0);
17 int a[maxn],b[maxn],c[maxn],d[maxn];
18 int main(int argc,const char *argv[]){
19     string str1,str2,str3,str4;
20     int len1,len2,len3,len4;
21     int up,n;
22     cin>>n;
23     while(n--){
24         cin>>str1>>str2>>str3>>str4;
25         len1=str1.length();len2=str2.length();
26         len3=str3.length();len4=str4.length();
27         memset(a,0,sizeof(a));memset(b,0,sizeof(b));
28         int i,j,k;
29         for(i=len1-1,k=0;i!=-1;--i){
30             a[k]=str1[i]-‘0‘;
31             k++;
32         }
33         for(j=len2-1,k=0;j!=-1;--j){
34             b[k]=str2[j]-‘0‘;
35             k++;
36         }
37         for(i=0,up=0;i<25;++i){
38             a[i]=a[i]+b[i]+up;
39             up=a[i]/10;
40             a[i]%=10;
41         }
42         memset(b,0,sizeof(b));memset(c,0,sizeof(c));
43         for(i=len3-1,k=0;i!=-1;--i){
44             b[k]=str3[i]-‘0‘;
45             k++;
46         }
47         for(j=len4-1,k=0;j!=-1;--j){
48             c[k]=str4[j]-‘0‘;
49             k++;
50         }
51         for(int i=0,up=0;i<maxn;++i){
52             b[i]=b[i]+c[i]+up;
53             up=b[i]/10;
54             b[i]%=10;
55         }
56         for(i=0,up=0;i<maxn;++i){
57             a[i]=a[i]+b[i]+up;
58             up=a[i]/10;
59             a[i]%=10;
60         }
61         for(i=maxn-1;i>=0;i--){
62             if(a[i])break;
63         }
64         if(i==-1)
65             cout<<"0";
66         else
67             for(;i>=0;i--)
68                 cout<<a[i];
69         cout<<endl;
70     }
71     return 0;
72 }

 

 

 

Rabbits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1193    Accepted Submission(s): 628

Problem Description

Here N (N ≥ 3) rabbits are playing by the river. They are playing on a number line, each occupying a different integer. In a single move, one of the outer rabbits jumps into a space between any other two. At no point may two rabbits occupy the same position.
Help them play as long as possible

 

 

Input

The input has several test cases. The first line of input contains an integer t (1 ≤ t ≤ 500) indicating the number of test cases.
For each case the first line contains the integer N (3 ≤ N ≤ 500) described as above. The second line contains n integers a1 < a2 < a3 < ... < aN which are the initial positions of the rabbits. For each rabbit, its initial position
ai satisfies 1 ≤ ai ≤ 10000.

 

 

Output

For each case, output the largest number of moves the rabbits can make.

 

 

Sample Input

5

3

3 4 6

3

2 3 5

3

3 5 9

4

1 2 3 4

4

1 2 4 5

 

 

Sample Output

1

1

3

0

1

 

 

Source

2017ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

 

 

 

代码:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=1e5+10;
 4 const int INF=0x3f3f3f3f;
 5 int flag[N];
 6 int a[N];
 7 int main(){
 8     int t,n,minn;
 9     int num1,num2,num;
10     while(~scanf("%d",&t)){
11         while(t--){
12            scanf("%d",&n);
13            memset(flag,0,sizeof(flag));
14            for(int i=0;i<n;i++){
15                 scanf("%d",&a[i]);
16                 flag[a[i]]=1;
17            }
18            num1=0,num2=0,num=0;
19            for(int i=a[0];i<=a[1];i++){
20                 if(flag[i]==0)num1++;
21            }
22            for(int i=a[n-2];i<=a[n-1];i++){
23                 if(flag[i]==0)num2++;
24            }
25            for(int i=a[0];i<=a[n-1];i++){
26                 if(flag[i]==0)num++;
27            }
28            //cout<<num1<<" "<<num2<<" "<<num<<endl;
29            minn=min(num1,num2);
30            printf("%d\n",num-minn);
31         }
32     }
33     return 0;
34 }