RE 0分 求调

常规
21

我不会打

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22

我不会用cout<<a+b<<endl;

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23

endl我都没学过

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24

@lhd endl不打也行

2 个赞
25

题面没作要求

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26

我不太会用加法

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27

AC代码

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int a,b;
    cin >> a >> b;
    cout << a+b << endl;
    return 0;
}
4 个赞
28

GDB告诉我cur->nxt->pre = cur;RE了，是因为cur->nxt是空指针吗？

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29

这题无视 Problem这个东西

4 个赞
30

直接printf(“%d”,a+b);

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31

不可以对head 或 tail 进行 merge 或 split

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32

怎么可以这么粗暴地刷解决方案

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33

！！！

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34

但本来这样就解决了呀

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35

感觉你这是个天才

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36

并且t1和t2初始不在链表中

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37

难道你是真的天才

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38

感觉调了以后还是RE

#include<bits/stdc++.h>
using namespace std;
const int k = 700;
const int _nk = 1000;
const int _2k = 1400;
struct node;
using np = node*;
struct node {
	vector<int>*a;
	np pre, nxt;
};
np _Pl;
np _Pr;
queue<np>bin;
int _Cnt = 32;
inline void CreateNewPlace() {
	_Pr = (_Pl = (np)(malloc(sizeof(node) * _Cnt))) + _Cnt;
	_Cnt <<= 1;
}
inline np newnp() {
	if (bin.size()) {
		np tmp = bin.front();
		bin.pop();
		return tmp;
	}
	if (_Pl == _Pr)CreateNewPlace();
	return (_Pl++);
}
inline void delnp(np x) {
	bin.push(x);
	x->a->clear();
	x->a = nullptr;
	x->pre = x->nxt = nullptr;
}
inline void checks(np cur);
inline void checkm(np cur);
inline void split(np cur) {
	if(cur ->pre ==cur || cur->nxt == cur || cur == nullptr)return;
	np tmp = newnp();
	int ts = cur->a->size() >> 1;
	vector<int> tv(cur->a->begin() + ts, cur->a->end());
	tmp->a = &tv;
	cur->a->erase(cur->a->begin() + ts, cur->a->end());
	tmp->nxt = cur->nxt;
	tmp->pre = cur;
	tmp->nxt->pre = tmp;
	tmp->pre->nxt = cur;
	checks(tmp);
	checks(cur);
	return;
}
inline void merge(np cur, np tmp) {
	if(cur ->pre == cur || tmp ->nxt == tmp || cur == tmp)return;
	cur->a->insert(cur->a->end(), tmp->a->begin(), tmp->a->end());
	cur->nxt = tmp->nxt;
	cur->nxt->pre = cur;
	delnp(tmp);
	checkm(cur);
	return;
}
inline void checks(np cur) {
	if (cur->a->size() >= _2k)split(cur);
	return;
}
inline void checkm(np cur) {
	if ((cur->a->size() + cur->nxt->a->size()) < _nk)merge(cur, cur->nxt);
	if ((cur->pre->a->size() + cur->a->size()) < _nk)merge(cur->pre, cur);
}
node head, tail;
inline void init() {
	head.pre = &head;
	head.nxt = &tail;
	tail.pre = &head;
	tail.nxt = &tail;
}
int main() {
	init();
	int a, b;
	cin >> a >> b;
	vector<int>veca(1, a);
	vector<int>vecb(1, b);
	node t1, t2;
	t1.nxt = &t2;
	t2.pre = &t1;
	t1.pre = &head;
	t2.nxt = &tail;
	head.nxt = &t1;
	tail.pre = &t2;
	t1.a = &veca;	
	t2.a = &vecb;
	merge(&t1,&t2);
	split(&t1);
	cout << *(head.nxt->a->begin()) + *(tail.pre->a->begin());		
	return 0;
}
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39

此帖子已被社区举报，现已被临时隐藏。

41

警告一次，不要回老帖

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