Codeforces Round #629 (Div. 3) A、B、C

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A：Divisibility Problem

大意：T组数据给定a b ，a每次只能加一，问多少次操作后能让a%b==0

思路：如果a比b大，那么答案是(a/b+1)*b-a或者直接输出0（不用操作）

如果a比b小，答案是b-a

代码：

#include<bits/stdc++.h> 
using namespace std;
#define LL long long
#define INF 2000000000
#define eps 1e-8
#define pi  3.141592653589793
const LL mod = 1e9+7;
using namespace std;
int main() {
    LL a,b;
    int _;
    for(scanf("%d",&_);_--;){
        scanf("%lld %lld",&a,&b);
        if(a<b){
            printf("%d
",b-a);
        }else{
            if(a%b == 0){
                puts("0");
            }else{
                printf("%lld
",(a/b+1)*b-a);
            }
        }
        
    }
}/*
 
*/

B：K-th Beautiful String

For the given integer

Recall that the string

For example, if

aaabb
aabab
aabba
abaab
ababa
abbaa
baaab
baaba
babaa
bbaaa

It is easy to show that such a list of strings will contain exactly

You are given

Input

The input contains one or more test cases.

The first line contains one integer

Each test case is written on the the separate line containing two integers

The sum of values

Output

For each test case print the

Example

input

Copy

output

Copy

aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa

大意：

需要输出的字符串中有且必有2个b，其余都是a。

T组数据，给定n和k，询问有n个字母构成的字符串第k大的字符串是啥

思路：

观察到abb 是第1个 bab是第2个 baab是第4个 baaab是第7个，再手动计算了一下baaaab是第11个。

由此很容易的看得出来如果一个b在末尾，前一个b移动的规律是数组[1,2,3,4....]的前缀和。

也很好理解，因为每次都要加上后面一个b的活动范围。

那么就很好算得出第一个b的位置了，直接打表二分即可。算出第一个b的位置，第二个b也好算啦。

代码：

#include<bits/stdc++.h> 
using namespace std;
#define LL long long
#define INF 2000000000
#define eps 1e-8
#define pi  3.141592653589793
const LL mod = 1e9+7;
using namespace std;
vector<LL>v;
int main() {
    LL ans = 1;
    for(int i = 2 ; i <= 100000 ; i ++){
        v.push_back(ans);
        ans = ans+(i-1);
    }
    LL a,b;
    int _;
    for(scanf("%d",&_);_--;){
        scanf("%lld %lld",&a,&b);
        int pos1 = lower_bound(v.begin(),v.end(),b)-v.begin(),pos2;//pos1是第一个b的位置，二分算得，pos2是第二个b的位置，由输入的k和pos1共同决定
        if(b == v[pos1]){
            pos2 = 0;
            pos1 += 1;
        }else{
            pos2 = b-v[pos1-1];
        }
        string s = "";
        for(int i = 0 ; i < a ; i ++){
            s += 'a';
        }
        s[a-pos2-1] = 'b';
        s[a-pos1-1] = 'b';
        cout<<s<<endl;
    }
}/*
bb    1
bab   2
baab  4
baaab 7
 
*/

C：Ternary XOR

A number is ternary if it contains only digits

You are given a long ternary number

Let's define the ternary XOR operation

Your task is to find such ternary numbers

You have to answer

Input

The first line of the input contains one integer

Output

For each test case, print the answer — two ternary integers

Example

input

Copy

output

Copy

思路：对0，1，2的情况分类讨论即可。

代码：

#include<bits/stdc++.h> 
using namespace std;
#define LL long long
#define INF 2000000000
#define eps 1e-8
#define pi  3.141592653589793
const LL mod = 1e9+7;
using namespace std;
int main() {
    string s;
    int _;
    for(scanf("%d",&_);_--;){
        cin>>s>>s;
        string s1 = "",s2 = "";
        int flag = 0;
        for(int i = 0 ; i < s.size() ; i ++){
            if(s[i] == '2'){
                if(flag == 0){
                    s1 += '1';
                    s2 += '1';
                }else{
                    s1 += '0';
                    s2 += '2';
                }
            }else if(s[i] == '1'){
                if(flag == 0){
                    s1 += '1';
                    s2 += '0';
                    flag = 1;
                }else{
                    s1 += '0';
                    s2 += '1';
                } 
            }else if(s[i] == '0'){
                s1 += '0';
                s2 += '0';
            }
        }
        cout<<s1<<endl;
        cout<<s2<<endl;
    }
}/*
4
5
22222
5
21211
1
2
9
220222021
 
 
*/

Codeforces Round #629 (Div. 3) A、B、C

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