如何从递归代码中删除重复项?许多6面骰子将被卷起,并且只打印一些独特的组合,这些组合加起来就达到了所需的总和。

基于@jarod42评论的示例性解决方案:

void rep(int n, std::string ans, int currentSum, int TargetSum, int startFrom) {
   if (n == 0) {
      if (currentSum == TargetSum)
         std::cout << "{" << ans << "}" << std::endl;
   } else 
      for (int i = startFrom; i <= 6; i++) 
         if (n > 1)
            rep(n-1, ans + to_string(i) + ", ", currentSum + i, TargetSum, i);
         else
            rep(n-1, ans + to_string(i), currentSum + i, TargetSum, i);
}

void ManyDiceSum(int howManyDice, int TargetSum) {
   if (howManyDice > 0)
      rep(howManyDice, "", 0, TargetSum, 1);
}  

首先,您不需要将目标和当前总和都传递给函数,只有目标是足够的,首先检查它将显著加快整个过程:

void rep( int n, int target, std::string ans = std::string(), int last = 1 )
{
    if( n == 0 ) {
        if( target == 0 )
            std::cout << "{" << ans << "}" << std::endl;
        return;
    }
    const auto limit = std::min( 6, target );
    // or even following, but not sure this is quite right though
    //const auto limit = std::min( 6, target - n * last ); 
    for( int i = last ; i <= limit; ++i ) 
        rep(n-1, target - i, ans + to_string(i) + ( n > 1 ? ", " : "" ), i );
}

void ManyDiceSum( int howManyDice, int TargetSum ) 
{
   if (howManyDice > 0)
      rep( howManyDice, TargetSum );
}  

为了避免重复,请从上次启动的号码继续。