Codeforces Round #707 (Div. 2)A~C题解
写在前边
链接：Codeforces Round #707 (Div. 2)
心态真的越来越不好了，看A没看懂，赛后模拟了一遍就过了，B很简单，但是漏了个判断重复的条件。
A. Alexey and Train
链接：A题链接
题目大意:
不想说了，题目看了半天没看懂，心态又看炸了。
思路：
代码:
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <map>
#include <cstring>
//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
using namespace std;
#define Inf 0x3f3f3f3f
#define PII pair<int, int>
#define P2LL pair<long long, long long>
#define endl '\n'
#define pub push_back
#define pob pop_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<long long> VLL;
typedef vector<int> VI;
const int Mod = 1000000007;
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
const int N = 110;
int a[N], b[N], tm[N];
void solve() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i] >> b[i];
}
for (int i = 1; i <= n; i++) {
cin >> tm[i];
}
int moment = 0;
for (int i = 1; i <= n; i++) {
moment = moment + a[i] - b[i - 1] + tm[i];
if (i == n) break;
int wait = (b[i] - a[i] + 1) / 2;
moment += wait;
if (moment >= b[i]) continue;
else moment = b[i];
}
cout << moment << endl;
}
int main()
{
//ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
int t;
scanf("%d", &t);
while (t--) {
solve();
}
return 0;
}
B. Napoleon Cake
链接：B题链接
题目大意:
就是往n层蛋糕上涂奶油，看最后有哪些层被奶油浸透。
思路：
- 双指针。
倒序枚举，枚举到一个涂有奶油的层，那么比它小的\(i - a[i] + 1\)都会被浸透,同时要注意如果遇到一个奶油更多的应该更新一下，比如1 0 0 0 4 3这个数据，枚举到3的时候，我们知道它能将蛋糕变成1 0 0 1 1 1，但是由于它的前边还有一个更厚的奶油，会使得蛋糕变成1 1 1 1 1 1所以应该要判断一下，详细说不清，看代码吧，主要要判断是否越界！
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <map>
#include <cstring>
//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
using namespace std;
#define Inf 0x3f3f3f3f
#define PII pair<int, int>
#define P2LL pair<long long, long long>
#define endl '\n'
#define pub push_back
#define pob pop_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<long long> VLL;
typedef vector<int> VI;
const int Mod = 1000000007;
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
const int N = 2e5 + 10;
int a[N];
bool st[N];
void solve() {
memset(st, false, sizeof(st));
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
for (int i = n; i >= 1; i--) {
if (a[i] > 0) {
int j = i;
int temp = a[i];
while (temp) {
st[j] = true;
temp--;
j--;
if (j <= 0) break;
if (a[j] >= temp) break;
}
i = j + 1;
}
}
for (int i = 1; i <= n; i++) {
if (st[i]) printf("%d ", 1);
else printf("%d ", 0);
}
cout << endl;
}
int main()
{
//ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
int t;
scanf("%d", &t);
while (t--) {
solve();
}
return 0;
}
2.差分
让\(b[i + 1]--\), 让\(b[max(i - a[i] + 1, 1)]++\) 也挺巧妙。
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <map>
#include <cstring>
//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
using namespace std;
#define Inf 0x3f3f3f3f
#define PII pair<int, int>
#define P2LL pair<long long, long long>
#define endl '\n'
#define pub push_back
#define pob pop_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<long long> VLL;
typedef vector<int> VI;
const int Mod = 1000000007;
const int N = 2E5 + 10;
int b[N], n, a[N];
void solve() {
memset(b, 0, sizeof(b));
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
for (int i = 1; i <= n; i++) {
b[i + 1]--;
b[max(i - a[i] + 1, 1)]++;
}
for (int i = 1; i <= n; i++) b[i] += b[i - 1];
for (int i = 1; i <= n; i++) printf("%d ", (b[i] > 1 ? 1 : b[i]));
puts("");
}
int main()
{
//ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
int t;
scanf("%d", &t);
while (t--) {
solve();
}
return 0;
}
C. Going Home
链接：C题链接
题目大意:
给定\(n\)个数，找出四个坐标\(x，y，z，w\)使得\(a_x + a_y = a_z + z_w\)
思路：
这个题是真的没想到，暴力\(O(n^2)\)随便过，原理是抽屉原理，比如有8个苹果7个箱子，现在将苹果全部装入箱子，那么至少有一个箱子有两个苹果，而现在给了N个数，那么我们可以有\(N * (N - 1) / 2\)个对，如果\(\cfrac{N * (N - 1)}{2} > 5,000,000\)，那么其中必然有和相同的对。一共可以枚举出\(\cfrac{n*(n - 1)}{2}\)个数对，但是由于两个数最大之和为\(5*10^6\)，我们所能枚举到的和的个数最多也就是\(5*10^6\)，我们可以把先枚举到的和存到一个数组里，而最多枚举到\(5*10^6\)个数我们就能接着枚举到一个重复的和，而大部分情况下根本不用枚举到\(5*10^6\)个数我们就能枚举到一个已经存在的数，那么跳出循环即可,所以复杂度就是\(O(min(n^2, n + c))\)，做过好几个类似的题了，都是可以通过判断出最大枚举范围然后直接暴力来做的。
代码
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <map>
#include <cstring>
//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
using namespace std;
#define Inf 0x3f3f3f3f
#define PII pair<int, int>
#define P2LL pair<long long, long long>
#define endl '\n'
#define pub push_back
#define pob pop_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<long long> VLL;
typedef vector<int> VI;
const int Mod = 1000000007;
const int N = 2e5 + 10, M = 5e6 + 10;
int a[N];
PII temp[M];
void solve() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
for (int i = 1; i <= n - 1; i++) {
for (int j = i + 1; j <= n; j++) {
int t = a[i] + a[j];
if (temp[t].first == 0 || temp[t].second == 0) {
temp[t] = {i, j};
continue;
} else {
if (temp[t].first != i && temp[t].first != j && temp[t].second != i && temp[t].second != j) {
puts("YES");
printf("%d %d %d %d\n", temp[t].first, temp[t].second, i, j);
return;
}
}
}
}
puts("NO");
}
int main()
{
//ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
solve();
return 0;
}
