Codeforces Round #640 (Div. 4)
作者：互联网
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...
标签：typedef,Created,int,ll,Codeforces,long,640,Div,include 来源： https://www.cnblogs.com/Vampire6/p/12862488.html