docs(B4050): 添加 GESP202409

高精算法---1.求n!的值/1.求n!的值.md

@@ -0,0 +1,53 @@
+## 1、求N！的值 【问题描述】        
+用高精度方法，求N！的精确值(N以一般整数输入)。 
+【输入样例】ni.in 
+```in
+10
+```
+【输出样例】ni.out
+```out
+3628800
+```
+
+```cpp
+#include<iostream>
+#include<string>
+#include<vector>
+using namespace std;
+
+// 高精度字符串乘法：res * num
+string multiply(string res, int num) {
+    vector<int> temp;
+    int carry = 0;
+    for (int i = res.size() - 1; i >= 0; i--) {
+        int product = (res[i] - '0') * num + carry;
+        temp.push_back(product % 10);
+        carry = product / 10;
+    }
+    while (carry) {
+        temp.push_back(carry % 10);
+        carry /= 10;
+    }
+
+    // 把结果翻转回来
+    string result;
+    for (int i = temp.size() - 1; i >= 0; i--) {
+        result += temp[i] + '0';
+    }
+    return result;
+}
+
+int main() {
+    int n;
+    cin >> n;
+    string result = "1";
+
+    for (int i = 2; i <= n; i++) {
+        result = multiply(result, i);
+    }
+    reverse(result.begin(), result.end());
+    cout << result << endl;
+    return 0;
+}
+
+```

高精算法---1.求n!的值/ni.cpp

@@ -0,0 +1,33 @@
+#include<iostream>
+#include<cstring>
+using namespace std;
+int a[100000],n,i,y,xy[100000];
+
+int main()
+{
+	cin>>n;
+	a[0]=1;
+	a[1]=1;
+	for (y=1;y<=n;y++) 
+	{
+		memset(xy,0,sizeof(xy));
+	    xy[0]=a[0];
+	    for (i=1;i<=a[0];i++)
+	    {
+		  xy[i]+=a[i]*y;
+		  xy[i+1]=xy[i]/10;
+		  xy[i]%=10;
+		}
+	    while (xy[xy[0]+1]>0) 
+		{
+			xy[xy[0]+2]=xy[xy[0]+1]/10;
+			xy[xy[0]+1]%=10;
+			xy[0]++;
+	    }
+	    for (i=1;i<=xy[0];i++) a[i]=xy[i];
+	    a[0]=xy[0];
+	}
+	for (i=a[0];i>=1;i--) cout<<a[i];
+	cout<<endl;
+	return 0;
+}

高精算法---1.求n!的值/ni.pas

@@ -0,0 +1,34 @@
+var
+  i,j,n,w : longint;
+  a : array[1..1000] of integer;
+
+procedure fact(k : longint);
+var
+  x,i : longint;
+begin
+  x := 0;
+  for i := 1 to w do
+  begin
+    a[i] := a[i]*k+x;
+    x := a[i] div 10;
+    a[i] := a[i] mod 10;
+  end;
+  while x>0 DO
+  begin
+    w := w+1;
+    a[w] := x mod 10;
+    x := x div 10;
+  end;
+end;
+
+begin
+  assign(input,'ni.in');  reset(input);
+  assign(output,'ni.out');  rewrite(output);
+  a[1] := 1;
+  w := 1;
+  readln(n);
+  for i := 1 to n do fact(i);
+  for j := w downto 1 do write(a[j]);
+  writeln;
+  close(input);  close(output);
+end.

高精算法---1.求n!的值/ni_data/ni0.in

@@ -0,0 +1 @@
+10

高精算法---1.求n!的值/ni_data/ni0.out

@@ -0,0 +1 @@
+3628800

高精算法---1.求n!的值/ni_data/ni1.in

@@ -0,0 +1 @@
+20

高精算法---1.求n!的值/ni_data/ni1.out

@@ -0,0 +1 @@
+2432902008176640000

高精算法---1.求n!的值/ni_data/ni2.in

@@ -0,0 +1 @@
+55

高精算法---1.求n!的值/ni_data/ni2.out

@@ -0,0 +1 @@
+12696403353658275925965100847566516959580321051449436762275840000000000000

高精算法---1.求n!的值/ni_data/ni3.in

@@ -0,0 +1 @@
+88

高精算法---1.求n!的值/ni_data/ni3.out

@@ -0,0 +1 @@
+185482642257398439114796845645546284380220968949399346684421580986889562184028199319100141244804501828416633516851200000000000000000000

高精算法---1.求n!的值/ni_data/ni4.in

@@ -0,0 +1 @@
+133

高精算法---1.求n!的值/ni_data/ni4.out

@@ -0,0 +1 @@
+14872707060906857289084508911813048098675809251055070300508818286592035566485075754388082124671571841702793317081960037166525246368924700537538282948117301741317436012998958826217903503076596121600000000000000000000000000000000

高精算法---1.求n!的值/ni_data/ni5.in

@@ -0,0 +1 @@
+345

高精算法---1.求n!的值/ni_data/ni5.out

@@ -0,0 +1 @@
+24215638650792346558700053691985855570120556040258652734839783267039961720178323593174739047913617079695531502689473012213820889134885853992818438056445080201482863675240494802269823110125881000284687377104376400792200165127855908498047507347955446603093964326987087311394274684237308398502911304969719715098068025497504900730580217016573270011698467378924291550780873605154736879542602554635558428265690302091342359471863508627516511203478353542187151045838267239168928747525890559708487655213488727530884968558716385000436989129479527833010340517760688345368715729020015336862534353876914871201776699205878662858555857265544230999178449256448000000000000000000000000000000000000000000000000000000000000000000000000000000000000