didhv 题解分享 · 2024/4/4
八次求和(编程题) - 题解
``` import java.util.Scanner; public class Main { // 主函数 public static void main(String[] args) { // 读取输入的整数 Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); // 计算结果 long result = 0; for (int i = 1; i <= n; i++) { result += pow(i, 8); // 调用自定义幂函数并累加结果 result %= 123456789; // 取模操作 } // 输出结果 System.out.println(result); // 打印最终结果 } // 自定义幂函数 public static long pow(long base, long exponent) { long result = 1; while (exponent > 0) { if (exponent % 2 == 1) { result = (result * base) % 123456789; // 计算幂的过程中取模 } base = (base * base) % 123456789; // 计算幂的过程中取模 exponent >>= 1; // 右移一位,相当于除以2 } return result; } } ```
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最后的路标 题解分享 · 2024/4/8
八次求和(编程题) - 题解
``` #include <bits/stdc++.h> #define objl '\n' typedef long long ll; using namespace std; //数据结构在这里定义 ll n; ll m = 123456789; ll bpm(ll b, ll p, ll m){ ll res = 1 % m; while(p!=0){ if(p&1){ res = (res % m) * (b % m) % m; } b = (b % m) * (b % m) % m; p = p >> 1; } return res; } void solve(){ //cin cout在这里 cin >> n; ll ans = 0; for(int x = 1; x<= n; x++){ ans = ((ans % m) + (bpm(x,8,m) % m)) % m; } cout << ans << endl; } int main(){ ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); int t;//多组数据要cin //cin >> t; t = 1; while(t--){ solve(); } return 0;//必须加return 0 } ```
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kaisyuantseng 题解分享 · 2024/4/11
八次求和(编程题) - 题解
```cpp #include <bits/stdc++.h> using namespace std; #define int long long // #define LL long long #define endl '\n' const int MOD = 123456789; int n; int quick_power(int a, int n) { a = a % MOD; int r = 1; while (n != 0) { if (n % 2 == 1) r = r * a % MOD; a = (a * a) % MOD; n = n / 2; } return r; } void solve() { cin >> n; int sum = 0; for (int i = 1; i <= n; i++) { sum += quick_power(i, 8); } cout << sum % MOD; } signed main() { ios::sync_with_stdio(0); cin.tie(0), cout.tie(0); solve(); return 0; } ```
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kzy 题解分享 · 2024/4/10
八次求和(编程题) - 题解
import java.math.BigInteger; import java.util.Scanner; public class Main { ``` public static void main(String[] args) { Scanner cin = new Scanner(System.in); int n = cin.nextInt(); BigInteger bi = new BigInteger("0"); for (int i = 0; i <= n; i++) { BigInteger temp = new BigInteger(i+""); temp = temp.pow(8); bi = bi.add(temp); bi = bi.mod(new BigInteger("123456789")); } System.out.println(bi.toString()); } ``` }
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Heng_Xin 题解分享 · 2024/4/7
八次求和(编程题) - 题解
```C++ #include <iostream> using namespace std; using ll = long long; ll mod = 123456789; ll myPow(ll a, ll b) { ll res = 1; while (b) { if (b & 1) res = res * a % mod; b >>= 1; a = a * a % mod; } return res; } int main() { int n; cin >> n; ll res = 0; for (int i = 1; i <= n; ++i) { res = (res + myPow(i, 8)) % mod; } cout << res << '\n'; return 0; } ```
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