Codeforces Round #701 (Div. 2)

Codeforces Round #701 (Div. 2)之心态爆炸

B - Replace and Keep Sorted

题意：
思路：因为都是前一个和后一个有关系，就得先记录贡献才行，没考虑到问的(l = r)，被×祭。
代码：

#include <iostream>
 
using namespace std;
typedef long long ll;
const ll maxn = 1e9 + 7;
const double eps = 1e-6;
const ll N = 1e5 + 999;
ll a[N];
ll sum[N];
void solve() {
	ll n, q, k;cin >> n >> q >> k;
	for (ll i = 1; i <= n; i++) 
		cin >> a[i];
	
	for (ll i = 2; i < n; i++) {
		sum[i] = sum[i-1]  - a[i-1] - 1 + a[i + 1]  - 1;
	}
	while (q--) {
		ll l, r;cin >> l >> r;
		ll ans = sum[r-1] - sum[l];
		if (r == l)ans = k-1;
		else 
		ans += -1 + a[l + 1]-1 + k -a[r-1]-1;
		cout << ans << endl;
	}
}
signed main() {
	ll t = 1;//cin >> t;
	while (t--)
	solve();	
}

D - Multiples and Power Differences

题意：
思路：猜到了是一列对角线是(lcm(1, 2, 3, cdots , 16)-x^4),一列是(lcm(1, 2, 3, cdots , 16))，然后因为循环时吧(k)写成(i)当场身亡。
代码：

#include <iostream>
 
using namespace std;
typedef long long ll;
const ll maxn = 1e9 + 7;
const double eps = 1e-6;
const ll N = 700;
ll a[N][N];
ll b[N][N];
ll f[167];
const ll NN = 720720;
void solve() {
	ll n, m;cin >> n >> m;
	for (ll i = 1; i <= n; i++) {
		for (ll j = 1; j <= m; j++) {
			cin >> a[i][j];
		}
	}
	for (ll  i = 1;i <= n; i++) {
		for (ll j = 1; j <= m; j++) {
			if ((i + j) % 2 == 1) {
				b[i][j] = NN;
			}else {
				for (ll k = 1; k <= 16; k++) {//k能写成i我是没想到的
					ll x = (k * k * k * k);
					if ( (NN-x) % a[i][j] == 0 && NN-x > 0) {
						b[i][j] = NN-x;
						break;
					}
				}
			}
			cout << b[i][j] << " ";
		}
		cout << endl;
	}
}
signed main() {
	ll t = 1;//cin >> t;
	while (t--)
	solve();	
}

总结：

1. 注意在前缀和时，用到了类似前缀和的数组，要判断询问是否(l = r)，很重要。

2. 注意在写代码时，不要用重复的变量名字。