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TangDou/AcWing/HighPrecision/高精度模板.md

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+### 一、算法原理
+
+1. 当输入的数很大时，可采用字符串方式接收。
+
+2. 拆成一位一位的数字，把它们存在一个数组中，一个数组元素表示一位数字
+
+数组中是这样存储的：
+<center><img src='https://img2020.cnblogs.com/blog/8562/202109/8562-20210917094727724-1157818874.png'></center>
+
+**倒序存储原因**：
+
+在平常，数字从左到右依次为从高位到低位....可这里却与日常的习惯相反。
+
+这是因为加法可能会产生进位，而数组在最前面加上数字是不可能的，但在尾巴处加上数字是好做的，所以，倒着放。
+
+<center> <img src='https://img2020.cnblogs.com/blog/8562/202109/8562-20210917095858016-1722478475.png'></center>
+
+
+### 二、[$AcWing$ $791$. 高精度加法](https://www.acwing.com/problem/content/793/)
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 100010;
+
+int a[N], b[N];
+int al, bl;
+void add(int a[], int &al, int b[], int &bl) {
+    int t = 0;
+    al = max(al, bl);
+    for (int i = 0; i < al; i++) {
+        t += a[i] + b[i];
+        a[i] = t % 10;
+        t /= 10;
+    }
+    if (t) a[al] = 1;
+    while (al > 0 && a[al] == 0) al--;
+}
+
+int main() {
+    string x, y;
+    cin >> x >> y;
+    for (int i = x.size() - 1; i >= 0; i--) a[al++] = x[i] - '0';
+    for (int i = y.size() - 1; i >= 0; i--) b[bl++] = y[i] - '0';
+    add(a, al, b, bl);
+    for (int i = al; i >= 0; i--) printf("%d", a[i]);
+    return 0;
+}
+```
+
+### 三、[$AcWing$ $792$. 高精度减法](https://www.acwing.com/problem/content/794/)
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+
+const int N = 1e5 + 10;
+int a[N], al;
+int b[N], bl;
+
+void sub(int a[], int &al, int b[], int &bl) {
+    int t = 0;
+    al = max(al, bl);
+    for (int i = 0; i < al; i++) {
+        t = a[i] - b[i] - t;
+        a[i] = (t + 10) % 10;
+        t < 0 ? t = 1 : t = 0;
+    }
+    while (al > 0 && a[al] == 0) al--;
+}
+
+int main() {
+    string x, y;
+    cin >> x >> y;
+
+    // 负号
+    if (x.size() < y.size() || (x.size() == y.size() && x < y)) {
+        printf("-");
+        swap(x, y);
+    }
+
+    for (int i = x.size() - 1; i >= 0; i--) a[al++] = x[i] - '0';
+    for (int i = y.size() - 1; i >= 0; i--) b[bl++] = y[i] - '0';
+
+    sub(a, al, b, bl);
+
+    for (int i = al; i >= 0; i--) printf("%d", a[i]);
+    return 0;
+}
+```
+
+### 四、[$AcWing$ $793$. 高精度乘法](https://www.acwing.com/problem/content/795/)
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 1e5 + 10;
+int a[N], al;
+int b[N], bl;
+void mul(int a[], int &al, int b[], int &bl) {
+    int t = 0;
+    int c[N] = {0};
+    for (int i = 0; i < al; i++)
+        for (int j = 0; j < bl; j++)
+            c[i + j] += a[i] * b[j];
+    al += bl;
+    for (int i = 0; i < al; i++) {
+        t += c[i];
+        a[i] = t % 10;
+        t /= 10;
+    }
+    // 前导0
+    while (al > 0 && a[al] == 0) al--;
+}
+
+int main() {
+    string x, y;
+    cin >> x >> y;
+    for (int i = x.size() - 1; i >= 0; i--) a[al++] = x[i] - '0';
+    for (int i = y.size() - 1; i >= 0; i--) b[bl++] = y[i] - '0';
+
+    mul(a, al, b, bl);
+    for (int i = al; i >= 0; i--) printf("%d", a[i]);
+    return 0;
+}
+
+```
+
+> **注**：网上对于高精度乘法，一般有高精乘低精 和 高精乘高精两种，学生们背起来太麻烦，容易出错，我把两个合并了一下，形成了一个模板，对于高精乘低精也可以兼容到，比如：
+```cpp {.line-numbers}
+int main() {
+    string x;
+    int y;
+    cin >> x >> y;
+    for (int i = x.size() - 1; i >= 0; i--) a[al++] = x[i] - '0';
+    b[0] = y, bl = 1;
+    mul(a, al, b, bl);
+    for (int i = al; i >= 0; i--) printf("%d", a[i]);
+    return 0;
+}
+```
+
+### 五、[$AcWing$ $794$. 高精度除法](https://www.acwing.com/problem/content/796/)
+
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+
+const int N = 1e5 + 10;
+int a[N], al, r;
+void div(int a[], int &al, int b, int &r) {
+    r = 0;
+    for (int i = al - 1; i >= 0; i--) {
+        r = r * 10 + a[i];
+        a[i] = r / b;
+        r %= b;
+    }
+    // 去前导0
+    while (al > 0 && !a[al]) al--;
+}
+
+int main() {
+    string x;
+    int y;
+
+    cin >> x >> y;
+
+    for (int i = x.size() - 1; i >= 0; i--) a[al++] = x[i] - '0';
+    // 高精度除法
+    div(a, al, y, r);
+
+    // 输出
+    for (int i = al; i >= 0; i--) printf("%d", a[i]);
+    puts("");
+    printf("%d\n", r);
+    return 0;
+}
+```
+
+### 六、保留结果$K$位
+```cpp {.line-numbers}
+// 高精乘高精,限长K(K<=100,不是很大)
+void mul(int a[], int b[]) {
+    int t = 0;
+    int c[N] = {0};
+    for (int i = 0; i < K; i++)
+        for (int j = 0; j < K; j++)
+            if (i + j < K) c[i + j] += a[i] * b[j]; // 防止越界，需要加上i+j<K的限制
+
+    for (int i = 0; i < K; i++) {
+        t += c[i];     // 进位与当前位置的和
+        a[i] = t % 10; // 余数保留下来
+        t /= 10;       // 进位
+    }
+}
+
+// 去前导0
+//int al=K;
+//while (al > 0 && a[al]==0) al--;
+```