2021牛客OI赛前集训营-普及组（第七场）(A+B+C+D)

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本文介绍了三道编程竞赛题目，涉及数组操作、图论和动态规划。在A题中，主要考察整数除法的巧妙应用；B题是一个关于边权值的调整问题，通过排序和更新求解；C题则是判断数组是否有序的简单问题。这些题目锻炼了参赛者的算法思维和编程技巧。

周末没事干了打算水一局普及局没想到翻跟头了

ll都可以忘了开！！！该死

【免费】2021牛客OI赛前集训营-普及组（第七场）_CSP/NOI/信息学奥赛培训提高_牛客竞赛OJ

A

int n, t;
int main()
{
    cin >> n >> t;
    cout << t % (n * n) / n + 1 << " " << t % n + 1;
    return 0;
}

B

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
template <typename T> void read(T &x) {
	x=0;
	char c=getchar();
	for (; !isdigit(c); c=getchar());
	for (; isdigit(c); c=getchar()) x=x*10+c-'0';
}
int n,k;
struct node {
	int u,v;
} a[1000005];
bool cmp(node a,node b) {
	return a.u<b.u;
}
int main() {
//	read(n),read(k);
	cin>>n>>k;
	for (int i=1; i<=n; ++i) {
//		read(a[i].v),read(a[i].u);
		cin>>a[i].v>>a[i].u;
	}
	sort(a+1,a+1+n,cmp);
	for (int i=1; i<=n; ++i) {
		if (a[i].v>a[i].u && k>=a[i].u) k=k-a[i].u+a[i].v;
	}
	printf("%d",k);
	return 0;
}

c

const int MAX=1e6+10;
int a[MAX];
int b[MAX];
int main(){
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i];
        b[i]=a[i];
    }
    sort(b,b+n,greater<int>());
    int flag=1;
    for(int i=0;i<n;i++){
        if(a[i]!=b[i]){
            flag=0;
            break;
        }
    }
    if(flag){
        cout<<0<<endl;
    }
    else {
        cout<<1<<endl;
    }
}

D

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

typedef long long LL;
const int N = 1e4 + 10;
const LL INF = 0x7f7f7f7f;

LL n, m, t, k, maxx;
LL v[60], w[60], a[N], dp[210];
LL dp1[45][45];

LL max(LL c, LL d)
{
    return c > d ? c : d;
}

LL min(LL c, LL d)
{
	return c > d ? d : c;
}

int main()
{
	cin >> n >> m >> t;
	for (LL i = 1; i <= n; i ++ )
	{
		memset(dp, 0, sizeof(dp));
		
		cin >> k;
		for (LL j = 1; j <= k; j ++ )
			cin >> w[j] >> v[j];
		
		for (LL j = 1; j <= k; j ++ )
        	for (LL k = t; k >= v[j]; k -- )
            	dp[k] = max(dp[k], dp[k - v[j]] + w[j]);

    	a[i] = dp[t];
	}
	
	for (int i = 1; i <= n - m + 1; i ++ )
	{
        memset(dp1, -INF, sizeof(dp1));
        
        for (int j = i; j < i + m; j ++ ) 
			dp1[j - i + 1][j - i + 1] = a[j];
			
        for (int len = 1; len < m; len ++ )
            for (int j = 1; j + len <= m; j ++ )
			{
                const int k = j + len;
                dp1[j][k] = max(dp1[j + 1][k] * 2ll + a[j + i - 1], dp1[j][k - 1] * 2ll + a[k + i - 1]);
            }
            
        maxx = max(maxx, dp1[1][m]);
    }
    
	cout << maxx << endl;
	return 0;
}