def x(i):
return 2**i
str = input()
str = str.replace("2(","x(")
print(eval(str))
#include #define ll long long
#define sc(a) scanf("%d", &a)
#define sc2(a, b) scanf("%d%d", &a, &b)
#define sc3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define scl(a) scanf("%lld", &a)
#define scl2(a, b) scanf("%lld%lld", &a, &b)
#define ss(a) scanf("%s", a)
#define mem(a, b) memset(a, b, sizeof(a))
#define pii pair#define ll __int128
using namespace std;
const ll mod = 998244353;
inline void print(__int128 x)
if (x > 9)
print(x / 10);
putchar(x % 10 + '0');
}template inline void read(_tp &x)
inline ll pow(ll a, ll b)
return ans;
}mapnum1; //存放x質因子和需要冪的數ai
mapnum2; //存放y質因子和需要冪的數bi
mapans; //共同質因子和需要冪的數
int main()
if (x != 1)
num1[x] = 1;
for (ll i = 2; i * i <= y; i++) //y質因子分解
if (y % i == 0)
if (y != 1)
num2[y] = 1;
for (auto p : num1)
}ll res = 1;
for (auto p : ans) //所有共同質因子冪次累乘
res = res * pow(p.first, p.second) % mod;
print(res % mod);
system("pause");
return 0;
}
#include #define ll long long
#define sc(a) scanf("%d", &a)
#define sc2(a, b) scanf("%d%d", &a, &b)
#define sc3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define scl(a) scanf("%lld", &a)
#define scl2(a, b) scanf("%lld%lld", &a, &b)
#define ss(a) scanf("%s", a)
#define mem(a, b) memset(a, b, sizeof(a))
#define pii pairusing namespace std;
const double pi = acos(-1.0);
const int mod = 1e9 + 7;
const int maxn = 1e6+5;
const int inf = 0x3f3f3f3f;
struct node
v.push_back(mii);//每天取最便宜的
}sort(v.begin(),v.end());//從小到大排
cout<#define ll long long
#define sc(a) scanf("%d", &a)
#define sc2(a, b) scanf("%d%d", &a, &b)
#define sc3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define scl(a) scanf("%lld", &a)
#define scl2(a, b) scanf("%lld%lld", &a, &b)
#define ss(a) scanf("%s", a)
#define mem(a, b) memset(a, b, sizeof(a))
#define pii pairusing namespace std;
const int maxn = 1e5+5;
int cnt[maxn],a[maxn];
int main()
int bas=0,ca=0;
for(int i=1;i<10;i++)
}for(int i=1;i<10;i++)
}for(int i=0;i<10;i++)}}
string b;
for(int i=ca;i>=1;i--)
if(t)b+=t+'0'; //最後進製
reverse(b.begin(),b.end());//字串反轉
cout<#define ll long long
#define sc(a) scanf("%d", &a)
#define sc2(a, b) scanf("%d%d", &a, &b)
#define sc3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define scl(a) scanf("%lld", &a)
#define scl2(a, b) scanf("%lld%lld", &a, &b)
#define ss(a) scanf("%s", a)
#define mem(a, b) memset(a, b, sizeof(a))
#define pii pairusing namespace std;
const int maxn = 1e5+5;
vectorv[maxn];
int pre[maxn],dep[maxn],maxd[maxn];
void add(int a,int b)
void dfs(int s,int fa)
}int main()
dep[n]=0;//根節點深度為0
dfs(n,-1);
int p=1;
while(t--)
//dep[p]就是第一種情況
//(maxd[p]+1)/2就是第二種情況
cout
return 0;
}
2019牛客多校第九場
由題意可設x y kp bx y kp b x y kp b代入第二個式子中可以得到kpx bx x2 c mod kpx bx x 2 equiv c mod kpx bx x2 c modp p p 第一項是p的倍數可以約掉,所以有x2 bx c 0 m od x 2 bx c equiv 0 ...
2020牛客多校第七場 解題報告BDH
include define ll long long define sc a scanf d a define sc2 a,b scanf d d a,b define sc3 a,b,c scanf d d d a,b,c define scl a scanf lld a define scl2...
牛客多校第九場補題(待完善)
i the crime solving plan of groundhog 題目大意 給定一組由 0 9 組成的數,組成兩個數使乘積最小。解題思路 用陣列儲存 0 9 的個數,先從 1 9 選最小的作為其中乙個乘數,其餘的數組成所能表達的最小的數。難點應該是大數乘法 這個也不難 include in...