B. Quasi Binary
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
A number is calledquasibinaryif its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011— are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integern. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integern(1≤n≤106).
Output
In the first line print a single integerk— the minimum number of numbers in the representation of numbernas a sum of quasibinary numbers.
In the second line printknumbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equaln. Do not have to print the leading zeroes in the numbers. The order of numbers doesn’t matter. If there are multiple possible representations, you are allowed to print any of them.
Sample test(s)
input
9
output
91 1 1 1 1 1 1 1 1
input
32
output
310 11 11
思路：例如543=500+40+3 表示500最少需要5个数，，表示40最少需要4个数，表示3最少需要3个数。
最大的数字为5，就一定至少要用5个数来表示。
因为可以输出任一组符合要求的数 所以输出 111 111 111 110 100
用字符数组做 15ms
#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){char a[10],b[10][10];while(~scanf("%s",a)){memset(b,0,sizeof(b));int cnt=0;int len=strlen(a);for(int i=0;i<len;i++){int m=a[i]-'0';if(m>cnt) cnt=m;for(int j=0;j<m;j++)b[j][i]='1';}printf("%d\n",cnt);for(int i=0;i<cnt;i++){int flag=0;for(int j=0;j<len;j++)if(b[i][j]=='1') flag=1;else if(b[i][j]!='1'&&flag) b[i][j]='0'; ///所缺的0，不是前导0}for(int i=0;i<cnt;i++){for(int j=0;j<len;j++)if(b[i][j]=='0'||b[i][j]=='1') printf("%c",b[i][j]); ///除了1 0 以外的都不需要输出if(i==cnt-1) puts("");else putchar(' ');}}return 0;}
用int做 15ms
#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){ int n; int num[7]={1,10,100,1000,10000,100000,1000000}; while(~scanf("%d",&n)) {int a[7],b[10];memset(a,0,sizeof(a));memset(b,0,sizeof(b));///b记录要输出的每一组数int cnt=0;int ans=0;while(n){a[cnt++]=n%10;if(n%10>ans) ans=n%10;n/=10;}<span style="font-family: Arial, Helvetica, sans-serif;">///找出最大的数字</span>printf("%d\n",ans);for(int j=0;j<cnt;j++){for(int i=0;i<a[j];i++)b[i]+=num[j];///比较容易错…}for(int i=0;i<ans-1;i++)printf("%d ",b[i]);printf("%d\n",b[ans-1]); } return 0;}
可你仍然感谢天地和人世所带来的这些变化和发生。
