1. T1

3.用vector存储数据.

   #include<bits/stdc++.h>
using namespace std;
int main(){
	int n,m;
	cin>>n>>m;
    vector<vector<int>> a;
    for(int i=0;i<n;i++){
    	a.push_back({});
    for(int j=0;j<m;j++){
        int b;
        cin>>b;
        a[i].push_back(b);
    }
    }

    for(int i=0;i<m;i++){
        for(int j=0;j<n;j++){
            cout<<a[j][i]<<' ';
        }
        cout<<endl;
    }
    return 0;
}

T3

贪心:在时间内做报酬最高的. 用优先队列维护数据.

#include<bits/stdc++.h>
using namespace std;

int main() {
    int n, m;
    cin >> n >> m;
    vector<pair<int, int> > a;
    for (int i = 1; i <= n; i++) {
        int x, y;
        cin >> x >> y;
        a.push_back({x, y});

    }

    sort(a.begin(), a.end());

    priority_queue<int> q;
   
    int ans = 0,j=0;
    for (int i=1;i<=m;i++) {
    	while (j < n && a[j].first <= i)
		{
			q.push(a[j].second);
			j++;
		}
		if (q.size() != 0)
		{
			ans += q.top(); 
			q.pop();
  }
}

    cout << ans << endl;

    return 0;
}

T4

因为数据范围很小,所以可以直接暴力枚举

#include<bits/stdc++.h>
using namespace std;
int main(){
    long long a,b,x;
    cin>>x;
    for(long long i=-200;i<=200;i++){
        for(long long  j=-200;j<=200;j++){
            if(i*i*i*i*i-j*j*j*j*j==x) {
                cout<<i<<' '<<j;
                return 0;
            }
        }
    }
    return 0;
}

T5

质因数分解+二分0--mid--p^e^ 对于每一个质因子可以找到一个最小的N 最后取最大值 N！=p~1~^e1^+p~2~^e2^+……+p~n~^en^ e~i~=n/p~i~

#include <iostream>
#include <unordered_map>
#include <algorithm>
#include <cmath>

using namespace std;

unordered_map<long long, int> factorize(long long K) {
    unordered_map<long long, int> factors;
    while (K % 2 == 0) {
        factors[2]++;
        K /= 2;
    }
    for (long long i = 3; i * i <= K; i += 2) {
        while (K % i == 0) {
            factors[i]++;
            K /= i;
        }
    }
    if (K > 1) {
        factors[K]++;
    }
    return factors;
}

long long count_p_in_factorial(long long N, long long p) {
    long long count = 0;
    while (N > 0) {
        N /= p;
        count += N;
    }
    return count;
}

long long find_min_N_for_prime(long long p, long long a) {
    long long low = 0;
    long long high = p * a;
    long long answer = 0;
    while (low <= high) {
        long long mid = (low + high) / 2;
        if (count_p_in_factorial(mid, p) >= a) {
            answer = mid;
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }
    return answer;
}
long long find_min_N(long long K) {
    if (K == 1) return 1;
    auto factors = factorize(K);
    long long max_N = 0;
    for (const auto& entry : factors) {
        long long p = entry.first;
        long long a = entry.second;
        long long N_p = find_min_N_for_prime(p, a);
        if (N_p > max_N) {
            max_N = N_p;
        }
    }
    return max_N;
}

int main() {
    long long K;
    cin >> K;
    cout << find_min_N(K) << endl;
    return 0;
}

T6

BFS搜索所有路径寻找最大差值

#include<bits/stdc++.h>
using namespace std;

const int N = 2e5 + 5;
const int inf = 0x3f3f3f3f;

int n , m , a[N] , dp[N] , d[N];
vector<int> g[N];
queue<int> q;

int main(){
	cin >> n >> m;
	
	for(int i = 1 ; i <= n ; i ++){
		cin >> a[i];
	}
	for(int i = 1 ; i <= m ; i ++){
		int x , y;
		cin >> x >> y;
		g[x].push_back(y);
		d[y] ++;
	}
	for(int i = 1 ; i <= n ; i ++){
		if(d[i] == 0) q.push(i);
		dp[i] = INT_MAX;
	}
	int ans = INT_MIN;
	while(!q.empty()){
		int u = q.front();
		q.pop();
		ans = max(ans , a[u] - dp[u]);
		
		for(auto v : g[u]){
			dp[v] = min(dp[v] , min(dp[u] , a[u]));
			if(-- d[v] == 0) q.push(v);
		}
	}
	cout << ans;
	return 0;
}