PAT (Advanced Level) Practise 1120Friend Numbers (20)

1120. Friend Numbers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Two integers are called "friend numbers" if they share the same sum of their digits, and the sum is their "friend ID". For example, 123 and 51 are friend numbers since 1+2+3 = 5+1 = 6, and 6 is their friend ID. Given some numbers, you are supposed to count the number of different friend ID's among them. Note: a number is considered a friend of itself.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N. Then N positive integers are given in the next line, separated by spaces. All the numbers are less than 10⁴.

Output Specification:

For each case, print in the first line the number of different frind ID's among the given integers. Then in the second line, output the friend ID's in increasing order. The numbers must be separated by exactly one space and there must be no extra space at the end of the line.

Sample Input:

8123 899 51 998 27 33 36 12

Sample Output:

43 6 9 26

简单题，直接用字符串读入统计个数就行了。
#include<cmath>#include<queue>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define ms(x,y) memset(x,y,sizeof(x))#define rep(i,j,k) for(int i=j;i<=k;i++)#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])#define inone(x) scanf("%d",&x)#define intwo(x,y) scanf("%d%d",&x,&y)#define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)#define lson x<<1,l,mid#define rson x<<1|1,mid+1,rconst int N = 1e3 + 10;const int INF = 0x7FFFFFFF;int n, f[N];int a[N], c;char s[N];int main(){  inone(n);  rep(i, 1, n)  {    scanf("%s", s);    int t = 0;    for (int j = 0; s[j]; j++) t += s[j] - '0';    if (f[t]++) continue;    a[c++] = t;  }  sort(a, a + c);  printf("%d\n", c);  rep(i, 0, c - 1)  {    printf("%s%d", i ? " " : "", a[i]);  }  return 0;}