Codeforces Round #587 (Div. 3) D~E1

还是思维不够啊切的好慢还手残
https://codeforces.com/contest/1216

D. Swords

这题每个人拿的武器一样而且还告诉你一开始每堆都一样而且又一个堆没有碰过最后告诉你每堆省多少
要求输出最少人数那样每人拿武器一定最多
武器最多肯定是差的 gcd 人最少

$n * x - s u m = z * y$
$x = a [n]$
$z = g c d$

E1. Numerical Sequence (easy version)

先暴力预处理了
找到它在什么范围那直接-到我们可以直接输出的范围就好
字符串在一直重复最多也加不过1e5次 num字符串长度要开长点

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int maxn = 1e7 + 5;
const long long INF = 0x3f3f3f3f3f3f3f3f;
typedef long long ll;
typedef pair<int, int> P;
int n, m, k, cas;
 
int a[maxn];
//n*x-sum=z*y
//x=a[n]
//z=gcd
signed main() {
	int n;
	int sum = 0;
	cin >> n;
	for(int i = 1; i <= n; i ++) {
		scanf("%d", &a[i]);
		sum += a[i];
	}
	sort(a + 1, a + 1 + n);
	int gcd = a[2] - a[1];
	
	for(int i = 3; i <= n; i ++) {
		gcd = __gcd(gcd, a[i] - a[i - 1]);
	}
	int p = a[n];
	int ans = (p * n - sum)/gcd;
	cout << ans << " " << gcd << endl;
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int maxn = 5e4 + 5;
const long long INF = 0x3f3f3f3f3f3f3f3f;
typedef long long ll;
typedef pair<int, int> P;
int n, m, k, cas;
 
int len[maxn];
int pc[maxn];
int num[500000], t = 0;
signed main() {
	for(int i = 1; i < maxn; i ++) {
		len[i] = len[i - 1] + log10(i) + 1;
	}
	pc[0] = 1;
	for(int i = 1; i < maxn; i ++)
		pc[i] = pc[i - 1] + len[i - 1];
	for(int i = 1; i < maxn; i ++) {
		int d = i, k = 0;
		char s[15];
		while(d) {
			s[k ++] = d % 10;
			d /= 10;
		}
		for(int j = k - 1; j >= 0; j --) num[++ t] = s[j];
	}
	int cas;
	cin >> cas;
	while(cas --) {
		cin >> n;
		for(int i = 1; i < maxn; i ++) {
			if(pc[i] <= n && pc[i + 1] > n) {
				cout << num[n - pc[i] + 1] << endl;
				break;
			}
		}
	}
	return 0;
}