Codeforces Round #609 (Div. 2) 【A,B,C】

 

 题意:给一个n<=1e7,找两个合数a和b使得a-b的差为n。

构造a=3n,b=2n,必含有公因子n,只有当n是1的时候是特例。

 1 #include<bits/stdc++.h>
 2  
 3 using namespace std;
 4 #define int long long
 5 #define inf 0x3f3f3f3f3f3f
 6 #define N 300009
 7 int arr[]={2,3,5,7,13};
 8 signed main(){
 9     int n;scanf("%lld",&n);
10     if(n==1){
11         cout<<"9 8";
12         return 0;
13     }
14     cout<<9*n<<" "<<8*n;
15     return 0;
16 }

 

 

 题意:给一个序列a,一个序列b,把这两个序列任意排序,然后使得对应位置的差在模意义下相等。

思路:排序+变相的尺取

 1 #include<bits/stdc++.h>
 2  
 3 using namespace std;
 4 #define int long long
 5 #define inf 0x3f3f3f3f3f3f3f3f3f3f3f3f3f
 6 #define N 200090
 7 int a[N],b[N],c[N];
 8 int n,m;
 9 int ok(int k){
10     for(int i=1;i<=n;i++){
11         c[i]=(k+a[i])%m; 
12     }
13     sort(c+1,c+1+n);
14     int f=1;
15     for(int i=1;i<=n;i++){
16         if(b[i]!=c[i]){
17             f=0;
18             return f;
19         }
20     } 
21     return f;
22 }
23 signed main(){
24     cin>>n>>m;
25     for(int i=1;i<=n;i++) cin>>a[i];
26     for(int i=1;i<=n;i++) cin>>b[i];
27     sort(b+1,b+1+n);
28     int minx=inf;
29     for(int i=1;i<=n;i++){
30         int temp=0;
31         if(a[i]>b[1]){
32             temp=m-(a[i]-b[1]);
33         }else{
34             temp=abs(a[i]-b[1]);
35         }
36         if(ok(temp)){
37             minx=min(minx,temp);
38         }
39     }
40     cout<<minx;
41     return 0;
42 } 

 

 

 

 题意:给一个n位10进制数字串s(首位不为0),构造一个数字串t(首位不为0),使得t串是有周期k,且t串>=s串,且t串最小。

直接按题意模拟。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define int long long
 4 #define N 300500
 5 int ans[N],arr[N],Temp[N];
 6 signed main(){
 7     int n,m;
 8     cin>>n>>m;
 9     string str;
10     cin>>str;
11     for(int i=0;i<str.size();i++) arr[i]=(str[i]-'0');
12     int f=1;
13     for(int i=m;i<n;i++){
14         if((str[i%m]-'0')>(str[i]-'0')){
15             break;
16         }
17         if((str[i%m]-'0')<(str[i]-'0')){
18             f=0;
19             break;
20         }
21     }
22     if(f){
23         cout<<n<<'\n';
24         for(int i=0;i<n;i++) cout<<(str[i%m]-'0');
25     }else{
26         int cnt=0;
27         for(int i=m-1;i>=0;i--){
28             Temp[cnt++]=(str[i]-'0');
29         }
30         Temp[0]=Temp[0]+1;
31         for(int i=0;i<=m;i++){
32             int x=Temp[i]%10;
33             int y=Temp[i]/10;
34             Temp[i]=x;
35             Temp[i+1]=Temp[i+1]+y;
36         }
37         for(int i=0;i<m;i++){
38             arr[i]=Temp[m-i-1];
39         }
40         cout<<n<<'\n';
41         for(int i=0;i<n;i++) cout<<arr[i%m];    
42     }
43     return 0;
44 }

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posted @ 2019-12-22 12:10  pengge666  阅读(262)  评论(0编辑  收藏  举报