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2025-08-30 18:35:01 +08:00
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19
TangDou/LuoGuBook/第8章模拟与高精度/P1009_1.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
//本题有坑，使用长整也过不了洛谷的用例，因为它的用例是22，
//需要高精度，也就是乘法高精度+加法高精度
typedef long long LL;
int n;
LL res;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) { // 1~n
LL f = 1; // 基底
for (int j = 1; j <= i; j++) f *= j; // f =n!
res += f; // 1!+2!+3!+...+n!
}
cout << res << endl;
return 0;
}

24
TangDou/LuoGuBook/第8章模拟与高精度/P1009_2.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
//高精度+高精度
//高精度*低精度
int main() {
int n;
cin >> n;
LL res = 0;
LL f = 1;
for (int i = 1; i <= n; i++) {
// f =i!
// 3!= 3*2*1
// 4!=4*3*2*1
f *= i; //计算i! =(i-1)! * i 举栗子： 4!= 4* 3!
res += f; //计算阶乘的和 res= 1!+2!+3!+...+n!
}
//输出要需要改
cout << res << endl;
return 0;
}

47
TangDou/LuoGuBook/第8章模拟与高精度/P1009_3.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
vector<int> add(vector<int> A, vector<int> B) {
if (A.size() < B.size()) swap(A, B);
vector<int> C;
int t = 0;
for (int i = 0; i < A.size(); i++) {
t += A[i];
if (i < B.size()) t += B[i];
C.push_back(t % 10);
t /= 10;
}
if (t) C.push_back(t);
return C;
}
vector<int> mul(vector<int> A, int b) {
vector<int> C;
int t = 0;
// 8899 * 2 = 17798
for (int i = 0; i < A.size() || t; i++) {
if (i < A.size()) t += A[i] * b;
C.push_back(t % 10);
t /= 10;
}
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
int main() {
int n;
cin >> n;
vector<int> F, RES;
//初始化
RES.push_back(0);
F.push_back(1);
for (int i = 1; i <= n; i++) { //
F = mul(F, i);
RES = add(RES, F);
}
//倒序输出
for (int i = RES.size() - 1; i >= 0; i--)
printf("%d", RES[i]);
return 0;
}

33
TangDou/LuoGuBook/第8章模拟与高精度/P1009_Prepare.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
//母题
//求n阶乘
LL f(int n) {
LL s = 1;
for (int i = 1; i <= n; i++) s *= i;
return s;
}
int main() {
//INT_MAX
//2147483647
// s=1!+2!+3!+...+n!
// 3!=3*2*1
// 4!=4*3*2*1 四步
// 4!=4*3! 一步，3！我知道
//20!=20*19*18*....*1
LL s = 0;
for (int i = 1; i <= 20; i++) s += f(i);
cout << s << endl;
LL t = 1;
s = 0;
for (int i = 1; i <= 20; i++) t = t * i, s += t;
cout << s << endl;
return 0;
}

18
TangDou/LuoGuBook/第8章模拟与高精度/P1009_Study.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n; // n!+(n-1)!+(n-2)!+...+1!
int sum = 0; //结果，累加和
for (int i = n; i >= 1; i--) { //枚举 n,n-1,n-2,....,1
//计算n这个数字的阶乘
int res = 1;
for (int j = i; j >= 1; j--) res *= j;
sum += res;
}
cout << sum << endl;
return 0;
}

48
TangDou/LuoGuBook/第8章模拟与高精度/P1042.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
const int N = 25 * 2500 + 10; //每行至多 25 个字母，最多有 2500 行。
int a[N];//小华的输赢状态表
int idx;
int main() {
char op;
while (true) {
cin >> op;
if (op == 'E') break;
if (op == 'W')
a[idx++] = 1;
else if (op == 'L')
a[idx++] = 0;
}
//开始分析11分
int w = 0, l = 0;
for (int i = 0; i < idx; i++) {
if (a[i])w++; else l++;
if ((w >= 11 || l >= 11) && abs(w - l) >= 2) {
printf("%d:%d\n", w, l);
//下一局开始，清零
w = l = 0;
}
}
//存在未打完的没分出胜负的
printf("%d:%d\n", w, l);
//两种计分制中间需要有空行
puts("");
//开始分析21分
w = 0, l = 0;
for (int i = 0; i < idx; i++) {
if (a[i])w++; else l++;
if ((w >= 21 || l >= 21) && abs(w - l) >= 2) {
printf("%d:%d\n", w, l);
//下一局开始，清零
w = l = 0;
}
}
printf("%d:%d\n", w, l);
return 0;
}

37
TangDou/LuoGuBook/第8章模拟与高精度/P1067.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
const int N = 110;
//这模拟太啰嗦了~
int main() {
int n;
cin >> n;
int x;
// a[0~n-1]是系数,a[n]是常数
for (int i = 0; i < n; i++) {
cin >> x;
if (x == 0) continue;
if (x < 0) {
if (x == -1)
cout << "-";
else
cout << x; //-1不要输出1，其它的-2，-3等如实输出
} else { // x肯定大于0
if (i > 0) cout << "+"; //第一位前面不能有+号
if (x != 1) cout << x; //系数1不输出，2，3等正常输出
}
if (n - i > 1) cout << "x^" << n - i;
else
cout << "x"; // 1次方特殊
}
//常数位
cin >> x;
if (x > 0)
cout << "+" << x;
else if (x < 0)
cout << x;
return 0;
}

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TangDou/LuoGuBook/第8章模拟与高精度/P1067_0.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/*
5
100 -1 1 -3 0 10
*/
int main() {
int n;
cin >> n;
int a;
for (int i = 0; i < n; i++) {
cin >> a;
if (a > 0) {
if (i > 0) cout << "+";
if (a > 1) {
if (n - i > 1)
cout << a << "x^" << n - i;
else
cout << a << "x";
}
else {
if (n - i > 1)
cout << "x^" << n - i;
else
cout << "x";
}
} else if (a < 0) {
if (a == -1) {
if (n - i > 1)
cout << "-x^" << n - i;
else
cout << "-x";
} else {
if (n - i > 1)
cout << a << "x^" << n - i;
else
cout << a << "x";
}
}
}
cin >> a;
if (a > 0)
cout << "+" << a << endl;
else if (a < 0)
cout << a << endl;
return 0;
}
/*
-x^99-7x^50+100x^2+x+8
-x^99-7x^50+100x^2+x^1+8
*/

48
TangDou/LuoGuBook/第8章模拟与高精度/P1067_1.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/*
5
100 -1 1 -3 0 10
*/
int main() {
int n;
cin >> n;
int a;
for (int i = 0; i < n; i++) {
cin >> a;
if (a > 0) {
if (i > 0) cout << "+";
if (a > 1) {
if (n - i == 1)
cout << a << "x";
else
cout << a << "x^" << n - i;
} else {
if (n - i == 1)
cout << "x";
else
cout << "x^" << n - i;
}
} else if (a < 0) {
if (a == -1) {
if (n - i == 1)
cout << "-x";
else
cout << "-x^" << n - i;
} else {
if (n - i == 1)
cout << a << "x";
else
cout << a << "x^" << n - i;
}
}
}
cin >> a;
if (a > 0)
cout << "+" << a << endl;
else
cout << a << endl;
return 0;
}

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TangDou/LuoGuBook/第8章模拟与高精度/P1067_2.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/*
5
100 -1 1 -3 0 10
*/
int main() {
int n;
cin >> n;
int a;
for (int i = 0; i < n; i++) {
cin >> a;
if (a > 0) {
if (i > 0) cout << "+";
if (a > 1) {
if (n - i == 1)
cout << a << "x";
else
cout << a << "x^" << n - i;
} else {
if (n - i == 1)
cout << "x";
else
cout << "x^" << n - i;
}
} else if (a < 0) {
if (a == -1) {
if (n - i == 1)
cout << "-x";
else
cout << "-x^" << n - i;
} else {
if (n - i == 1)
cout << a << "x";
else
cout << a << "x^" << n - i;
}
}
}
cin >> a;
if (a > 0)
cout << "+" << a << endl;
else if (a < 0)
cout << a << endl;
return 0;
}

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TangDou/LuoGuBook/第8章模拟与高精度/P1303.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/**
* 功能：高精度乘低精度模板
* @param A
* @param b
* @return
*/
vector<int> mul(vector<int> &A, int b) {
vector<int> C;
int t = 0;
for (int i = 0; i < A.size() || t; i++) {
if (i < A.size()) t += A[i] * b;
C.push_back(t % 10);
t /= 10;
}
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
int main() {
string a;
int b;
vector<int> A, C;
cin >> a >> b;
//倒着放的噢~
for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
C = mul(A, b);
//倒着输出噢~
for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
return 0;
}

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TangDou/LuoGuBook/第8章模拟与高精度/P1303_2.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/**
* 功能：高精度乘高精度模板
* @param A
* @param b
* @return
*/
vector<int> mul(vector<int> &A, vector<int> &B) {
//初始化大小
vector<int> C(A.size() + B.size());
//先放里再说
for (int i = 0; i < A.size(); i++)
for (int j = 0; j < B.size(); j++)
C[i + j] += A[i] * B[j];
//处理余数
for (int i = 0, t = 0; i < C.size(); i++) {
t += C[i];
if (i >= C.size()) C.push_back(t % 10);
else C[i] = t % 10;
t /= 10;
}
//去掉前导0
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
int main() {
string a, b;
cin >> a >> b;
//准备动作
vector<int> A, B;
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');
//计算
vector<int> C = mul(A, B);
//倒序输出
for (int i = C.size() - 1; i >= 0; i--) cout << C[i];
return 0;
}

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TangDou/LuoGuBook/第8章模拟与高精度/P1328.md Normal file
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## 问题
### 1、需要整理一个胜负的关系表
0 表示“剪刀”，1 表示“石头”，2 表示“布”，3 表示“蜥蜴人”，4表示“斯波克”。
前一个人的手势 后二个人的手势 前一个人的得分 数组表示
0 0 0 g[0][0]=0
0 1 0 g[0][1]=0
0 2 1 g[0][2]=1
0 3 1
0 4 0
1 0 1
1 1 0
1 2 0
1 3 1
1 4 0

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TangDou/LuoGuBook/第8章模拟与高精度/P1328_1.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
const int N = 210;
//0 表示“剪刀”，1 表示“石头”，2 表示“布”，3 表示“蜥蜴人”，4表示“斯波克”。
//看图打表写出对阵双方的胜负平结果二维数组
//这个胜负平表是关键
int g[5][5] = {
{0, 0, 1, 1, 0},
{1, 0, 0, 1, 0},
{0, 1, 0, 0, 1},
{0, 0, 1, 0, 1},
{1, 1, 0, 0, 0}
};
int n; //表示进行N次猜拳
int n1; //小A出拳的周期长度
int n2; //小B出拳的周期长度
int na[N]; //表示小A出拳的规律
int nb[N]; //表示小B出拳的规律
int s1; //小A赢几次
int s2; //小B赢几次
int main() {
//数据读入
cin >> n >> n1 >> n2;
for (int i = 0; i < n1; i++)cin >> na[i];
for (int i = 0; i < n2; i++)cin >> nb[i];
//开始猜拳
for (int i = 0; i < n; i++) {
//这一轮小A出的是啥？
int a = na[i % n1]; //n1为循环的长度，所以每一轮后，都需要mod n1
//这一轮小B出的是啥？
int b = nb[i % n2]; //n2为循环的长度，所以每一轮后，都需要mod n2
s1 += g[a][b]; //根据事先推出的胜负平表，找出胜负平关系，累加到小A的得分中。
s2 += g[b][a]; //根据事先推出的胜负平表，找出胜负平关系，累加到小A的得分中。
}
cout << s1 << " " << s2 << endl;
return 0;
}

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TangDou/LuoGuBook/第8章模拟与高精度/P1328_2.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
const int N = 210;
//本文件是示例数组下标从1开始如何计算
//数组下标从1开始时，需要注意的问题： 前n-1个用MOD没有问题，
// 关键是n的倍数时，需要输出n，而不是0，这个与下标从0开始时有区别！
//0 表示“剪刀”，1 表示“石头”，2 表示“布”，3 表示“蜥蜴人”，4表示“斯波克”。
//看图打表写出对阵双方的胜负平结果二维数组
//这个胜负平表是关键
int g[5][5] = {
{0, 0, 1, 1, 0},
{1, 0, 0, 1, 0},
{0, 1, 0, 0, 1},
{0, 0, 1, 0, 1},
{1, 1, 0, 0, 0}
};
int n; //表示进行N次猜拳
int n1; //小A出拳的周期长度
int n2; //小B出拳的周期长度
int na[N]; //表示小A出拳的规律
int nb[N]; //表示小B出拳的规律
int s1; //小A赢几次
int s2; //小B赢几次
int main() {
//数据读入
cin >> n >> n1 >> n2;
for (int i = 1; i <= n1; i++)cin >> na[i];
for (int i = 1; i <= n2; i++)cin >> nb[i];
//开始猜拳
for (int i = 1; i <= n; i++) {
//这一轮小A出的是啥？
int a = na[i % n1 == 0 ? n1 : i % n1];
//n1为循环的长度，所以每一轮后，都需要mod n1
//这一轮小B出的是啥？
int b = nb[i % n2 == 0 ? n2 : i % n2];
//n2为循环的长度，所以每一轮后，都需要mod n2
s1 += g[a][b]; //根据事先推出的胜负平表，找出胜负平关系，
// 累加到小A的得分中。
s2 += g[b][a]; //根据事先推出的胜负平表，找出胜负平关系，
// 累加到小A的得分中。
}
cout << s1 << " " << s2 << endl;
return 0;
}

35
TangDou/LuoGuBook/第8章模拟与高精度/P1563.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
int n, m;
const int N = 1e5 + 10;
struct Node {
int dir; //方向,0:圈内,1:圈外
char name[120]; //姓名
} c[N];
int main() {
//读入数据
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++)
scanf("%d%s", &c[i].dir, c[i].name);
//出发点
int i = 0;
while (m--) {
// a==0 ,表示向左数 s 个人
// a==1 ,表示向右数 s 个人
int a, s;
scanf("%d%d", &a, &s);
if (a == 0 && c[i].dir == 0) //圈内
i = (i - s + n) % n; //向前找
else if (a == 0 && c[i].dir == 1) //圈内
i = (i + s) % n; //向后找
else if (a == 1 && c[i].dir == 0) //圈外
i = (i + s) % n; //向后找
else if (a == 1 && c[i].dir == 1) //圈外
i = (i - s + n) % n; //向前找
}
printf("%s\n", c[i].name);
return 0;
}

47
TangDou/LuoGuBook/第8章模拟与高精度/P1601.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
/**
* 功能：高精度加法模板
* @param A
* @param B
* @return
*/
vector<int> add(vector<int> &A, vector<int> &B) {
//声明结果数组C
vector<int> C;
//以长的为根基，进行计算
if (A.size() < B.size()) return add(B, A);
//进位
int t = 0;
//列竖式进行计算
for (int i = 0; i < A.size(); i++) {
t += A[i];//进位加法
if (i < B.size()) t += B[i]; //存在这一位，就加
C.push_back(t % 10);//保留余数
t /= 10;//进位
}
//最后可能还有进位
if (t) C.push_back(t);
//返回
return C;
}
int main() {
//以字符串形式读入
string a, b;
cin >> a >> b;
//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
vector<int> A, B, C;
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
C = add(A, B);
//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;
}

33
TangDou/LuoGuBook/第8章模拟与高精度/P2670.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
const int N = 110;
char a[N][N];
int dx[] = {0, 0, -1, 1, -1, 1, -1, 1};
int dy[] = {-1, 1, 0, 0, -1, -1, 1, 1}; //左右上下 左上 左下 右上 右下
//
int n, m;
int main() {
scanf("%d%d", &n, &m);
//计入地雷地图
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
scanf("%d", &a[i][j]);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (a[i][j] == '*')
printf("%c", '*');
else {
int cnt = 0;
for (int k = 0; k < 8; k++)
//当前点周边的*的个数是多少个？
if (a[i + dx[k]][j + dy[k]] == '*') cnt++;
printf("%d", cnt);
}
}
printf("\n");
}
return 0;
}

42
TangDou/LuoGuBook/第8章模拟与高精度/P4924.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
const int N = 510;
int n, m;
int x, y, r, z;
int tmp[N][N], res[N][N];
int cnt;
// 数组res为最后输出的结果数组，tmp为辅助数组；
int main() {
cin >> n >> m;
//初始化数据
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
res[i][j] = tmp[i][j] = ++cnt;
//读入魔法指令
while (m--) { // m次指令
cin >> x >> y >> r >> z;
//处理旋转
if (z == 0) { //顺时针
for (int i = -r; i <= r; i++)
for (int j = -r; j <= r; j++)
tmp[x + j][y - i] = res[x + i][y + j];
} else if (z == 1) { //逆时针
for (int i = -r; i <= r; i++)
for (int j = -r; j <= r; j++)
tmp[x - j][y + i] = res[x + i][y + j];
}
//复制回来
memcpy(res, tmp, sizeof tmp);
}
//输出
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++)
printf("%2d ", res[i][j]);
printf("\n");
}
return 0;
}

60
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### P4924 [1007]魔法少女小Scarlet
#### 1、初始化数组的代码
```c++
int cnt = 0; //计数器
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
a[i][j] = ++cnt;
```
....
01 02 03 04 05 06 07 08 09
10 11 12 13 14 15 15 17 18
....
#### 2、%02d的格式控制
```c++
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++)
printf("%02d ", res[i][j]);
printf("\n");
}
```
#### 3、矩阵的旋转
我们模拟原位置(i,j)在第一象限，i>0,j>0
逆时针90度，到达的位置是: (-j,i)
顺时针90度，到达的位置是:(j,-i)
180度，到达的位置：(-i,-j)
想方法背下来：
(1)1，3象限，i,j位置不变。
2,4象限，i,j对调
(2)1象限是正的，3象限全是负的
2象限x是负的，y是正的
4象限x是正的，y是负的
#### 4、矩阵的遍历
```c++
//逆时针
for (int i = -r; i <= r; i++)
for (int j = -r; j <= r; j++)
f[x - j][y + i] = res[x + i][y + j];
//顺时针
for (int i = -r; i <= r; i++)
for (int j = -r; j <= r; j++)
f[x + j][y - i] = res[x + i][y + j];
```
#### 5、临时的矩阵变量
防止出现自行覆盖的问题，采用的办法是做一个数组的拷贝，复制品。
```c++
memcpy(res, f, sizeof f);
```

50
TangDou/LuoGuBook/第8章模拟与高精度/P4924_1.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
// Rotate Image（二维数组逆时针旋转90度）
const int N = 110;
int n = 9, cnt;
int res[N][N], f[N][N];
//矩阵旋转，需要额外开创建一个临时的二维数组(防止出现交换赋值时，出现覆盖的问题)
void rotate(int x, int y, int r) { // x:第几行 y:以(x,y)点为中心，四个方向，向左r,向右r，转~
for (int i = -r; i <= r; i++)
for (int j = -r; j <= r; j++) //
f[x - j][y + i] = res[x + i][y + j]; //副本修改
// 复制数组的正规写法
memcpy(res, f, sizeof f); //将副本复制回到正本上
//复制
// for (int i = 1; i <= n; i++)
// for (int j = 1; j <= n; j++)
// res[i][j] = f[i][j];
}
int main() {
//初始化填充数据
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
res[i][j] = f[i][j] = ++cnt;
//输出结果
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++)
printf("%02d ", res[i][j]);
printf("\n");
}
printf("\n");
//测试一下旋转
rotate(6, 7, 1);
//输出结果
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++)
printf("%02d ", res[i][j]);
printf("\n");
}
return 0;
}

50
TangDou/LuoGuBook/第8章模拟与高精度/P4924_2.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
// Rotate Image（二维数组顺时针旋转90度）
const int N = 110;
int n = 9, cnt;
int res[N][N], tmp[N][N];
//矩阵旋转，需要额外开创建一个临时的二维数组(防止出现交换赋值时，出现覆盖的问题)
void rotate(int x, int y, int r) {
for (int i = -r; i <= r; i++)
for (int j = -r; j <= r; j++)
//每一个位置都去了哪里
tmp[x + j][y - i] = res[x + i][y + j];
//复制回来
memcpy(res, tmp, sizeof tmp);
// for (int i = 1; i <= n; i++)
// for (int j = 1; j <= n; j++)
// res[i][j] = tmp[i][j];
}
int main() {
//初始化数据
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
res[i][j] = tmp[i][j] = ++cnt;
//输出结果
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++)
printf("%2d ", res[i][j]);
printf("\n");
}
printf("\n");
//测试一下旋转
rotate(6, 7, 1);
//输出结果
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++)
printf("%2d ", res[i][j]);
printf("\n");
}
return 0;
}

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TangDou/LuoGuBook/第8章模拟与高精度/TD.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
typedef long long LL;
/*
5
100 -1 1 -3 0 10
100x^5-x^4+x^3-3x^2+10
*/
int main() {
int n;
cin >> n;
for (int i = 1; i <= n + 1; i++) {
int a;
cin >> a;
if (a == 1) {
if (i == 1)
cout << "x^" << n - i + 1;
else
cout << "+x^" << n - i + 1;
}
else if (a == -1)
cout << "-"
<< "x^" << n - i + 1;
else if (a > 0) {
if (i == 1)
cout << a << "x^" << n - i + 1;
else
cout << "+" << a << "x^" << n - i + 1;
} else
cout << a << "x" << n - i + 1;
}
return 0;
}

3
TangDou/LuoGuBook/第8章模拟与高精度/Test.md Normal file
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$f(x)=2x^2+2x+4$
$f(x)=x^5+3x^2+2x^2-9$