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.
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.
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
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
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.
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
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 }