'commit'

TangDou/LuoGuBook/P2142.cpp

@@ -0,0 +1,49 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+/**
+ * 功能：高精度减法模板
+ * @param A
+ * @param B
+ * @return
+ */
+vector<int> sub(vector<int> &A, vector<int> &B) {
+    int t = 0;
+    vector<int> C;
+    for (int i = 0; i < A.size(); i++) {
+        t = A[i] - t;
+        if (i < B.size()) t -= B[i];
+        C.push_back((t + 10) % 10);
+        if (t < 0)t = 1; else t = 0;
+    }
+    while (C.size() > 1 && C.back() == 0) C.pop_back();
+    return C;
+}
+
+//对比函数
+bool cmp(vector<int> &A, vector<int> &B) {
+    if (A.size() != B.size()) return A.size() > B.size();
+    for (int i = A.size() - 1; i >= 0; i--)  //这里的判断要注意顺序，因为高位在后
+        if (A[i] != B[i]) return A[i] > B[i];
+    return true;
+}
+
+int main() {
+    //以字符串形式读入
+    string a, b;
+    cin >> a >> b;
+    vector<int> A, B, C;
+    //1、准备动作，将字符串的每一位，反序放入到整数运算数组中
+    for (int i = a.size() - 1; i >= 0; i--)A.push_back(a[i] - '0');
+    for (int i = b.size() - 1; i >= 0; i--)B.push_back(b[i] - '0');
+    //2、调用模板进行计算
+    if (cmp(A, B)) C = sub(A, B);
+    else {
+        printf("-");
+        C = sub(B, A);
+    }
+    //3、输出结果,需要倒序噢
+    for (int i = C.size() - 1; i >= 0; i--)printf("%d", C[i]);
+    return 0;
+}