归并 快排 快速幂

/*
归并排序：不断将数组切分为两部分
        递归切分直到数组长度为1，之后逐渐有序合并
*/
#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;

//归并 小-->大
void Merge(int a[], int s, int m, int e, int tmp[]) {
    int iter = s;
    int it1 = s, it2 = m+1;

    while(it1 <=m && it2 <= e) {
        if(a[it1] < a[it2]) tmp[iter++] = a[it1++];
        else tmp[iter++] = a[it2++];
    }
    while(it1 <= m)
        tmp[iter++] = a[it1++];
    while(it2 <= e)
        tmp[iter++] = a[it2++];
    for(int i=s; i<=e; ++i) {
        a[i] = tmp[i];
    }
    return;
}
void MergeSort(int a[], int s, int e, int tmp[]) {
    if(s < e) {
        int m = s + (e-s)/2;
        MergeSort(a, s, m, tmp);
        MergeSort(a, m+1, e, tmp);
        Merge(a, s, m, e, tmp);
    }
    return;
}
int a[] = {13, 12, 23, 5, 2, 34, 11, 34, 13};
int tmp[100];
int main() {
    int size = sizeof(a) / sizeof(int);
    MergeSort(a, 0, size-1, tmp);
    for(int i=0; i<size; ++i)
        printf("%d ", a[i]);
}

/*
    快排：主要思想，寻找切分元素之后两边分治处理
*/
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn = 1e5 + 10;
int num[maxn];
int n;
inline void quik_sort(int s, int e) 
{
    //mid为切分元素
    int mid = num[s+(e-s)/2];

    // i 左侧指针，找到第一个不小于切分元素mid的数，i指向>=mid
    // j 右侧指针，找到第一个不大于切分元素mid的数，j指向<=mid
    int i = s, j = e;
    while(i <= j) {
        while(num[i] < mid) i += 1;
        while(num[j] > mid) j -= 1;

        if(j >= i) {
            swap(num[i], num[j]);
            i += 1, j -= 1;
        }
    }
    //跳出后j < i, 分成两部分继续
    if(s < j) quik_sort(s, j);
    if(e > i) quik_sort(i, e);
    return;
}
int main() {
    scanf("%d", &n);
    for(int i=1; i<=n; ++i) 
        scanf("%d", num+i);
    quik_sort(1, n);
    for (int i = 1; i <= n; ++i)
    {
        printf("%d ", num[i]);        
    }
    return 0;
}

#include <cstdio>
#include <iostream>
using namespace std;
int Mod ;

//将指数二进制化，也可以尝试其他进制
int quick_pow(int base, int t) {
    int ans = 1;
    while(t) {
        if(t & 1) {
            ans = (ans%Mod)*(base%Mod);
　　　　　　　ans %= Mod;
        }
        base = (base*base) % Mod;
        t >>= 1;
    }
    return ans;
}
int main()
{
    int base, t;
    scanf("%d %d %d", &base, &t, &Mod);
    printf("%d^%d mod %d=%d\n", base, t, Mod, quick_pow(base, t));
    return 0;
}