高精度，50tps，tle，求调！

corner_xiejunqi @ 2024-08-27 10:40:47

#include<bits/stdc++.h>
using namespace std;
const int N=1e7+10;
int n;
int a[N]; 
int main(){
    //1、声明变量，输入
    cin>>n;
    //2、计算过程
    int k=1;
    memset(a,0,sizeof a);
    a[1]=1;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=k;j++){
            a[j]*=2;
        }
        int len=k;
        if(a[k]>=10){
            k++;
        }
        for(int j=1;j<=len;j++){
            if(a[j]>=10){
                a[j+1]+=a[j]/10;
                a[j]%=10;
            }
        }

    }
    //3、输出
    cout<<k;
    for(int i=500;i>=3;i--){
        if(i%50==0){
            cout<<endl;
        }
        cout<<a[i];
    }
    if(a[1]>=1){
        cout<<a[2]<<a[1]-1;
    }else{
        cout<<a[2]-1<<a[1]+10-1;
    }
    return 0;
}

by cgxd @ 2024-08-27 17:57:55

计算时可以优化一下，如果位数超过500，就按照500算，计算位数的时候可以改为

printf("%d", int(log(2) / log(10) * n) + 1);

by cgxd @ 2024-08-27 18:00:33

还有乘方是可以用快速幂

by cgxd @ 2024-08-27 18:01:36

可以用我的代码参考一下

#include<bits/stdc++.h>
using namespace std;
int n;
string ans;
unordered_map<string, unordered_map<string, string>> m;
string operator+(string s1, string s2){
    if(s1.size() < s2.size())
        swap(s1, s2);
    if(s2 == "0")
        return s1;
    reverse(s1.begin(), s1.end());
    reverse(s2.begin(), s2.end());
    s2.resize(s1.size(), '0');
    s1.push_back('0');
    for(size_t i = 0; i < s2.size(); ++i){
        s1[i] += s2[i] - '0';
        if(s1[i] > '9'){
            s1[i] -= 10;
            ++s1[i + 1];
        }
    }while(s1.back() == '0')
        s1.pop_back();
    if(s1.empty())
        return "0";
    if(s1.size() > 500)
        s1.resize(500);
    reverse(s1.begin(), s1.end());
    return s1;
}string operator*(string s1, string s2){
    if(s1 == "0" or s2 == "0")
        return "0";
    if(s2 == "10"){
        s1.push_back('0');
        return s1;
    }if(m.count(s1) and m[s1].count(s2))
        return m[s1][s2];
    if(s1.size() == 1 and s2.size() == 1)
        return m[s1][s2] = to_string((s1[0] - '0') * (s2[0] - '0'));
    string ans("0");
    for(char c: s1)
        ans = ans * "10" + s2 * string(1, c);
    return m[s1][s2] = ans;
}bool cmp(string s1, string s2){
    if(s1.size() != s2.size())
        return s1.size() > s2.size();
    return s1 > s2;
}string operator-(string s1, string s2){
    if(!cmp(s2, s1)){
        reverse(s1.begin(), s1.end());
        reverse(s2.begin(), s2.end());
        s2.resize(s1.size(), '0');
        s1.push_back('0');
        for(size_t i = 0; i < s2.size(); ++i){
            int k = s2[i] - '0';
            s1[i] -= k;
            if(s1[i] < '0'){
                s1[i] += 10;
                --s1[i + 1];
            }
        }while(s1.back() == '0')
            s1.pop_back();
        if(s1.empty())
            return "0";
        reverse(s1.begin(), s1.end());
        return s1;
    }else{
        string s3 = s2 - s1;
        auto it = s3.begin();
        s3.insert(it, '-');
        return s3;
    }
}string power(string s, int k){
    string ans = "1";
    do{
        if(k & 1)
            ans = ans * s;
        s = s * s;
    }while(k >>= 1);
    return ans;
}int main(){
    scanf("%d", &n);
    ans = power("2", n) - "1";
    printf("%d", int(log(2) / log(10) * n) + 1);
    reverse(ans.begin(), ans.end());
    ans.resize(500, '0');
    reverse(ans.begin(), ans.end());
    for(int i = 0; i < 500; ++i){
        if(i % 50 == 0)
            printf("\n");
        putchar(ans[i]);
    }return 0;
}

by corner_xiejunqi @ 2024-08-29 17:22:07

@cgxd 感谢