谁能给我解释一下这个CodeForces 919B/Perfect Number解决方案？

Question

这是我为CodeForces问题919B找到的一个解决方案--完美数。从技术上讲，我理解它在做什么，但我想理解它背后的‘直觉’或想法/方法。

int main(){
    int k=0, m=19, c=0, sum=0;
    scanf("%d", &k);
    while(true){
        int n = m;
        sum = 0;
        while(n){
            sum+=n%10;
            n=n/10;
        }
        printf("%d %d %d\n", n, sum, c);
        if(sum == 10) c++;
        if(c == k) break;
        m++;
    }
    printf("%d", m);
    return 0;
}

Answer 1

用户0x499602D2已经给出了一个非常简洁的答案，但让我详细说明一下。超高级伪代码将如下所示：

loop over natural numbers and check if they are perfect:
  if found k such numbers: 
    stop and output the last perfect number

下面是一个更详细的伪代码，它与您的c++代码片段相匹配：

m   = 19 # first perfect number (1+9=10) - a good starting point
c   = 0  # total number of perfect numbers found so far
sum = 0  # temporary variable that will hold the sum of digits

read k from stdin # we're going to look for the k-th perfect number 

# we start an open-ended search but we know that we will find k-th number and stop
for m = 19 ... infinity:
  if m is perfect:
    increment c by one        # found c perfect numbers so far

  if c == k:           
    exit loop and return m    # m is the k-th perfect number! we're done!

看看c与我们找到的“完美数”的数量是如何对应的？一旦c命中k，我们的循环就可以停止。

还有一个技术问题:我们如何检查数字1945是否完全正确？我们需要对数字求和: 1+9+4+5。从数字n中获得最低有效数字的简单方法是将除数的余数除以10 (即n modulo 10)：

1945 % 10 = 5
194  % 10 = 4
19   % 10 = 9
1    % 10 = 1

如何从1945到194再到19到1？只需做整数除以10：

1945/10 = 194
 194/10 = 19
  19/10 = 1
   1/10 = 0 -> stop the loop, since while(0) is the same as while(false)

这就是在printf之前发生的事情。