题目链接：http://codeforces.com/contest/1352

A

思路：9876拆成9000+800+70+6就行了，其他类似这样拆，注意当某一位为0就不用处理了。

//-------------------------------------------------
//Created by HanJinyu
//Created Time :一  5/ 4 16:56:31 2020
//File Name :632B.cpp
//-------------------------------------------------

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef double db;
typedef long long ll;
const int maxn = 200005;

int main()
{
    #ifdef ONLINE_JUDGE
    #else
    freopen("in.txt","r",stdin);
    #endif
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        string s;
        cin>>s;
        bool flag=false;
        int ops=0;
        for(int i=0;i<s.length();i++)
        {
            if(s[i]!='0'&&i!=0)
            {
                flag=true;
                ops++;
            }
            else if(i==0&&s[i]!='0'){
                    ops++;
            }
        }
        if(!flag){
            cout<<"1"<<endl<<s<<endl;
        }
        else
        {
            cout<<ops<<endl;
            for(int i=0;i<s.length();i++){
                if(s[i]!='0'){
                    cout<<s[i];
                    for(int j=0;j<s.length()-i-1;j++){
                        cout<<'0';
                    }
                    cout<<" ";
                }
            }
            cout<<endl;
        }
    }
     return 0;
}

View Code

B

思路：根据n和k的奇偶性判断，n是奇数，k是偶数是不能找到k个奇偶性相同的数之和为n的，k>n也是不满足的，然后就是奇奇，偶偶，偶奇的情况，奇奇的话全部一开始k个数全为1，然后剩余的偶数全部加在这k个数的任何一个数，还是奇数，偶偶的话判断2*k和n的大小，<的话一开始全置2，>的话一开始只能全置1，否则不够，剩余的还是加在k个数当中的任何一个即可，偶奇的话也要判断2*k和n的大小，<的话一开始全置2，剩余的再加到其中任何一个，>的话无法找到奇偶性相同的k个数之和为n。

//-------------------------------------------------
//Created by HanJinyu
//Created Time :一  5/ 4 16:56:31 2020
//File Name :632B.cpp
//-------------------------------------------------

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef double db;
typedef long long ll;
const int maxn = 200005;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("in.txt","r",stdin);
#endif
    int T;
    scanf("%d",&T);
    while(T--){
        ll n;
        int k;
        scanf("%lld%d",&n,&k);
        if(k>n||(n%2!=0&&k%2==0))
            printf("NO\n");
        else if(n%2==0&&k%2==0)
        {
            if(2*k<=n)
            {
                printf("YES\n");
                for(int i=0;i<k-1;i++)
                    printf("2 ");
                printf("%lld \n",n-2*(k-1));
            }
            else{
                printf("YES\n");
                for(int i=0;i<k-1;i++)
                    printf("1 ");
                printf("%lld \n",1+n-k);
            }

        }
        else if(n%2!=0&&k%2!=0){
            printf("YES\n");
            for(int i=0;i<k-1;i++)
            {
                printf("1 ");
            }
            printf("%lld \n",1+n-k);
        }
        else if(n%2==0&&k%2!=0){
            if(2*k<=n){
                printf("YES\n");
                for(int i=0;i<k-1;i++)
                    printf("2 ");
                printf("%lld\n",2+n-2*k);
            }
            else
            {
                printf("NO\n");
            }
        }
    }
    return 0;
}

View Code

C

思路：计算k/(n-1)他与n相乘加上k%(n-1)就行了，注意一下结果是否还能整除n,能的话要减一。

//-------------------------------------------------
//Created by HanJinyu
//Created Time :一  5/ 4 16:56:31 2020
//File Name :632B.cpp
//-------------------------------------------------

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef double db;
typedef long long ll;
const int maxn = 200005;

int main()
{
    #ifdef ONLINE_JUDGE
    #else
    freopen("in.txt","r",stdin);
    #endif
    int T;
    scanf("%d",&T);
    while(T--){
        ll n,k;
        cin>>n>>k;
        ll mm=k/(n-1);
        ll nn=k%(n-1);
        ll result=mm*n+nn;
        if(result%n==0)
            cout<<result-1<<endl;
        else
            cout<<result<<endl;
    }
     return 0;
}

View Code

D

待续...