這道題其實也不難,遞歸的想法很簡單:
把目前n個骰子分成兩堆:1堆1個(可以枚舉點數1-6),1堆有n-1個。遞推到最後,n=0時,累計起來的點數的次數就+1啦
之所以寫這篇博客,是因爲我覺得書中的遞歸代碼寫得有點混亂(並沒有貶低作者的意思),所以貼一下我自己認爲比較好的寫法:
遞歸解法
#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;
typedef unsigned int uInt;
const int low = 1, high = 6;
void prob(vector<int>& count, uInt n, int currentSum) {
if (n == 0) {
++count[currentSum];
return;
}
--n;
for (int i = low; i <= high; ++i) {
prob(count, n, currentSum+i);
}
}
void cal(uInt n) {
vector<int> count(high*n+1);
prob(count, n, 0);
int totalCount = pow(6, n), sum = 0;
for (uInt i = n; i < count.size(); ++i) {
sum += count[i];
printf("點數爲%d的概率爲%.3lf, count=%d\n",
i, count[i]*1.0/totalCount, count[i]);
}
printf("sum=%d, totalCount=%d\n\n", sum, totalCount);
}
int main() {
cal(1);
cal(2);
cal(3);
return 0;
}
輸出結果爲(其中有挺多調試信息的):
點數爲1的概率爲0.167, count=1
點數爲2的概率爲0.167, count=1
點數爲3的概率爲0.167, count=1
點數爲4的概率爲0.167, count=1
點數爲5的概率爲0.167, count=1
點數爲6的概率爲0.167, count=1
sum=6, totalCount=6
點數爲2的概率爲0.028, count=1
點數爲3的概率爲0.056, count=2
點數爲4的概率爲0.083, count=3
點數爲5的概率爲0.111, count=4
點數爲6的概率爲0.139, count=5
點數爲7的概率爲0.167, count=6
點數爲8的概率爲0.139, count=5
點數爲9的概率爲0.111, count=4
點數爲10的概率爲0.083, count=3
點數爲11的概率爲0.056, count=2
點數爲12的概率爲0.028, count=1
sum=36, totalCount=36
點數爲3的概率爲0.005, count=1
點數爲4的概率爲0.014, count=3
點數爲5的概率爲0.028, count=6
點數爲6的概率爲0.046, count=10
點數爲7的概率爲0.069, count=15
點數爲8的概率爲0.097, count=21
點數爲9的概率爲0.116, count=25
點數爲10的概率爲0.125, count=27
點數爲11的概率爲0.125, count=27
點數爲12的概率爲0.116, count=25
點數爲13的概率爲0.097, count=21
點數爲14的概率爲0.069, count=15
點數爲15的概率爲0.046, count=10
點數爲16的概率爲0.028, count=6
點數爲17的概率爲0.014, count=3
點數爲18的概率爲0.005, count=1
sum=216, totalCount=216
迭代解法
如果理解了上面的遞歸解法,其實轉成迭代也很簡單!
假設我們知道了前n-1個骰子構成的點數的情況,比如點數爲2有1中情況,點數爲3有3種情況……,那麼就可以很輕鬆推斷出前n個骰子構成的點數的情況了!就是分別給前面的每個點數加上1,2,3,4,5,6,構成新的點數的情況數。
原書用了兩個數組,我用了兩個map,空間的重用性不夠好,不過可讀性增加了一點:
#include <cstdio>
#include <vector>
#include <cmath>
#include <map>
using namespace std;
void cal(int n) {
if (n <= 0)
return;
static const int low = 1, high = 6;
map<int, int> count;
count[0] = 1;
for (int i = 1; i <= n; ++i) {
map<int, int> now;
map<int, int>::iterator it = count.begin();
while (it != count.end()) {
int number = it->first, cnt = it->second;
for (int j = low; j <= high; ++j) {
now[number + j] += cnt;
}
++it;
}
count = now;
}
int totalCount = pow(6, n), sum = 0;
map<int, int>::iterator it = count.begin();
while (it != count.end()) {
sum += it->second;
printf("點數爲%d的概率爲%.3lf, count=%d\n",
it->first, it->second*1.0/totalCount, it->second);
++it;
}
printf("sum=%d, totalCount=%d\n\n", sum, totalCount);
}
int main() {
cal(1);
cal(2);
cal(3);
return 0;
}