Nothing for Nothing（十一）_nothing for nothing什么意思-优快云博客

本文链接：https://blog.youkuaiyun.com/REfusing/article/details/96990375

本文提供了六道CodeForces平台上的算法题目的详细解答，包括数组操作、双端队列应用、矩阵填充、区间查询、图论验证及物品丢弃问题等。通过这些题目，读者可以了解到不同数据结构的应用技巧及算法优化方案。

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文章目录

题目1：Nick and Array

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
int a[maxn];
int main()
{
    int n;
    scanf("%d", &n);
    for(int i= 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        if(a[i] >= 0)a[i] = -(a[i]+1);
    }
    int mm = 10, id = n+1;
    if(n % 2 != 0)
    {
        for(int i = 1; i <= n; i++)
        {
            if(a[i] < mm)
            {
                mm = a[i];
                id = i;
            }
        }
    }
    for(int i = 1; i <= n; i++)
    {
        if(i != 1)printf(" ");
        if(i == id)
        {
            printf("%d", -(a[i]+1));
            continue;
        }
        printf("%d", a[i]);
    }

    printf("\n");
    return 0;
}

题目2：Valeriy and Deque

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn  = 1e5+10;
int a[maxn];
#define LL long long
deque<int>q;
struct Node{
       int A, B;
}vis[maxn], tmp;
int res[maxn];
int main(){
    int n, m;
    scanf("%d %d", &n, &m);
    int maxx = -1, id;
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
        q.push_back(a[i]);
        if(maxx < a[i]){
            maxx = a[i];
            id = i;
        }
    }
    for(int i = 1; i < id; i++){
        int t1 = q.front();
        q.pop_front();
        int t2 = q.front();
        q.pop_front();
        //cout<<t1<<" "<<t2<<endl;
        tmp.A = t1;
        tmp.B = t2;
        vis[i] = tmp;
        if(t1 > t2){
            q.push_front(t1);
            q.push_back(t2);
        }
        else {
            q.push_front(t2);
            q.push_back(t1);
        }
    }
    tmp.A = q.front();
    q.pop_front();
    tmp.B = q.front();
    vis[id] = tmp;
    int i = 0;
    while(q.size()){
        int t = q.front();
        res[i++] = t;
        q.pop_front();
    }
    LL t;
    for(int i = 1; i <= m; i++){
        scanf("%lld", &t);
        if(t > id){
            printf("%d %d\n",maxx, res[(t-id) % (n-1)]);
        }
        else printf("%d %d\n",vis[t].A,vis[t].B );
    }

return 0;
}

题目3： Tolik and His Uncle

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n/2; i++)
    {
        for(int j = 1; j <= m; j++)
        {
            printf("%d %d\n", i, j);
            printf("%d %d\n", n-i+1, m-j+1);
        }
    }

    if(n % 2 == 1)
    {
        for(int i = 1; i <= m/2; i++)
        {
            printf("%d %d\n", n/2+1, i );
            printf("%d %d\n", n/2+1, m-i+1);
        }
        if(m % 2 == 1)printf("%d %d\n", n/2+1, m/2+1);
    }
    return 0;
}

题目4：Candies!

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
int a[maxn];
int sum[maxn];
void init(int n){
     sum[0] = 0;
     for(int i = 1; i <= n; i++)sum[i] += sum[i-1]+a[i];

}
int main(){
    int n, m;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)scanf("%d", &a[i]);
    init(n);
    scanf("%d", &m);
    while(m--){
        int l, r;
        scanf("%d %d", &l, &r);
        int ans = sum[r] - sum[l-1];
        printf("%d\n", ans/10);
    }

return 0;
}

题目5：Add on a Tree

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
int degree[maxn];
int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1; i < n; i++){
        int u, v;
        scanf("%d %d", &u, &v);
        degree[u]++;
        degree[v]++;
    }
    for(int i = 1; i <= n; i++){
        if(degree[i] == 2){
           printf("NO\n");
           return 0;
        }
    }
    printf("YES\n");
return 0;
}

题目6： Tokitsukaze and Discard Items

代码：

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

const int maxn = 1e5+10;
#define LL long long
LL m[maxn];
int main(){
    LL n, t, k;
    scanf("%lld %lld %lld", &n, &t, &k);
    for(int i = 0; i < t; i++)scanf("%lld", &m[i]);

    LL cn = 0, ans = 0, i = 0;
    while(i < t){

        LL block = (m[i]-cn-1)/k;
        LL val = (block+1)*k;
        LL pos = upper_bound(m, m+t, val+cn)-lower_bound(m, m+t, m[i]);
        ans++;
        cn += pos;
        i += pos;
    }
    printf("%lld\n", ans);
return 0;
}