Codeforces Round #748 (Div. 3) 题解代码
作者：互联网
题目链接
A
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
#define x first
#define y second
typedef pair<int,int> PII;
const int N = 1e4 + 10;
int main()
{
int T;
cin >> T;
while(T--)
{
int mx = 0;
int a[3];
cin >> a[0] >> a[1] >> a[2];
mx = max(max(a[0], a[1]), a[2]);
int cnt = 0;
for(int i = 0; i < 3; i++) if(a[i] == mx) cnt++;
for(int i = 0; i < 3; i++)
{
printf("%d ", a[i] == mx && cnt == 1 ? 0 : mx + 1 - a[i]);
}
puts("");
}
return 0;
}
B
这题需要美剧后面的两位，看看是否可以%25==0整除，然后前面的数组随意
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
#define x first
#define y second
typedef pair<int,int> PII;
typedef long long LL;
const int N = 1e4 + 10, INF = 0x3f3f3f3f;
LL k, n;
int nums[29], cnt, mx;
//len已经选择的数字
int get(int arr[])
{
int res = 0;
int i = cnt - 1, j = 1;
while(i >= 0 && j >= 0)
{
// printf("arr[j]=%d ", arr[j]);
while(i >= 0 && nums[i] != arr[j]) i--, res++;
// printf("del=%d ", nums[i + 1]);
// puts("");
if(i >= 0 && nums[i] == arr[j])
{
i--, j--;
// printf("i=%d,j=%d\n", i, j);
}
}
if(j >= 0) return res = INF;
// printf("n=%lld, %d%d, %d\n", k, arr[0],arr[1], res);
return res;
}
int main()
{
// #ifdef ONLINE_JUDGE
// #else
// freopen("in.txt", "r", stdin);
// #endif
int T;
cin >> T;
while(T--)
{
cin >> n;
k = n;
cnt = 0;
while(n) nums[cnt++] = n % 10, n /= 10;
reverse(nums, nums + cnt);
// for(int i = 0; i < cnt; i++) printf("%d ", nums[i]);
// puts("");
int res = INF;
int arr[2] = {0, 0};
res = min(res, get(arr));
arr[0] = 2, arr[1] = 5;
res = min(res, get(arr));
arr[0] = 5, arr[1] = 0;
res = min(res, get(arr));
arr[0] = 7, arr[1] = 5;
res = min(res, get(arr));
printf("%d\n", res);
}
return 0;
}
C
二分即可
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N = 4e5 + 10;
typedef long long LL;
int n, k;
int a[N];
bool check(int mid)
{
LL sum = 0;
for(int i = mid; i < k; i++)
{
if(sum >= a[i]) return false;
sum += n - a[i];
}
return true;
}
int main()
{
// #ifdef ONLINE_JUDGE
// #else
// freopen("D.txt", "r", stdin);
// #endif
int T;
cin >> T;
while (T--)
{
cin >> n >> k;
for(int i = 0; i < k; i++) cin >> a[i];
sort(a, a + k);
int l = 0, r = k - 1;
while(l < r)
{
int mid = l + r >> 1;
if(check(mid)) r = mid;
else l = mid + 1;
}
cout << k - l << endl;
}
return 0;
}
D1
如果不是全部相等，则求一个最大公约数就可以了
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef pair<int, int> PII;
#define x first
#define y second
typedef long long LL;
const int N = 4e5 + 10, M = N + N;
int n;
int a[N];
int gcd(int a, int b)
{
return !b ? a : gcd(b, a % b);
}
int main()
{
// #ifdef ONLINE_JUDGE
// #else
// freopen("G.txt", "r", stdin);
// #endif
int T;
cin >> T;
while(T--)
{
scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%d", &a[i]);
sort(a, a + n);
n = unique(a, a + n) - a;
if(n == 1) puts("-1");
else {
int mx = a[1] - a[0];
for(int i = 2; i < n; i++)
mx = gcd(mx, a[i] - a[i - 1]);
cout << mx << endl;
}
}
return 0;
}
D2
如果都能变到某一个数，则必定某个差值的约数d1出现的次数>=n/2
#include <iostream>
#include <cstring>
#include <map>
#include <unordered_map>
using namespace std;
const int N = 100;
typedef pair<int,int> PII;
int n;
int a[N];
PII p[N];
map<int,int> S;
unordered_map<int,int> cnt;
void get(int x, int k)
{
for(int i = 1; i * i <= x; i++)
{
if(x % i == 0)
{
cnt[i] += k;
if(x / i != i) cnt[x / i] += k;
}
}
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("D.txt", "r", stdin);
#endif
int T;
cin >> T;
while (T--)
{
S.clear();
cin >> n;
int mx = 0;
for(int i = 0; i < n; i++) cin >> a[i], mx = max(mx, ++S[a[i]]);
if(mx >= n / 2) {
puts("-1");
continue;
}
int len = 0;
for(auto&[k, v]: S) p[len++] = {k, v};
int ans = 0;
for(int i = 0; i < len; i++)
{
cnt.clear();
for(int j = i + 1; j < len; j++)
{
get(p[j].first - p[i].first, p[j].second);
}
for(auto&[k, v]: cnt)
{
if(v + p[i].second >= n / 2)
ans = max(ans, k);
}
}
cout << ans << endl;
}
return 0;
}
E
从度数为1的点开始bfs，简单模拟一下就可以
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef pair<int, int> PII;
#define x first
#define y second
typedef long long LL;
const int N = 4e5 + 10, M = N + N;
int n, k;
int h[N], e[M], ne[M], idx;
int d[N], q[N], p[N];
bool st[N];
void add(int a, int b)
{
e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
int main()
{
// #ifdef ONLINE_JUDGE
// #else
// freopen("G.txt", "r", stdin);
// #endif
int T;
cin >> T;
while (T--)
{
cin >> n >> k;
idx = 0;
if(n == 1) {
puts("0");
continue;
}
memset(h, -1, (n + 1) * 4);
memset(d, 0, (n + 1) * 4);
for(int i = 0; i < n - 1; i ++)
{
int a, b;
cin >> a >> b;
add(a, b), add(b, a);
d[b]++, d[a]++;
}
int hh = 0, tt = -1;
int tot = 0;
for(int i = 1; i <= n; i++) if(d[i] == 1) q[++tt] = i;
while(k && tot < n) {
int cnt = 0;
while(hh <= tt)
{
//删除当前点
int t = q[hh++];
tot++;
for(int i = h[t]; ~i; i = ne[i])
{
int j = e[i];
if(--d[j] == 1) p[cnt++] = j;
}
}
hh = 0, tt = -1;
for(int i = 0; i < cnt; i++) q[++tt] = p[i];
k--;
}
printf("%d\n", n - tot);
}
return 0;
}
F
f[i][j][a][b]表示，考虑了前i位，选了j位红色的数字，且红色数字modA=a，mod B = b的方案集合
属性：是否可以到达这个状态，true表示可以，false表示不行
状态转移
第i+1位红色：f[i][j][a][b] -> f[i + 1][j + 1][(a * 10 + x) % A][b]
黑色: f[i][j][a][b] -> f[i + 1][j][a][(b * 10 + x) % b]
刚开始的状态是f[0][0][0][0] = true
要输出方案，所以记录每个状态由哪几个状态转移而来, 由状态转移可以发现，后一个状态只需要记录前一个的余数，以及前一个被染成啥颜色即可
这里提供两种不同的写法，一种是倒推，一种是开一个数组记录上一个状态
记录上一个状态
#include<iostream>
#include<cstring>
#include<algorithm>
#include<string>
using namespace std;
const int N = 45;
bool f[N][N][N][N];
pair<bool, int> pre[N][N][N][N];
int n, A, B;
char s[N];
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("in.txt", "r", stdin);
#endif
int T;
cin >> T;
while(T--)
{
cin >> n >> A >> B >> s;
for(int i = 0; i <= n; i++)
for(int j = 0; j <= n; j++)
for(int a = 0; a < A; a++)
for(int b = 0; b < B; b++)
f[i][j][a][b] = false;
f[0][0][0][0] = true;
for(int i = 0; i < n; i++)
for(int j = 0; j <= i; j++)
for(int a = 0; a < A; a++)
for(int b = 0; b < B; b++)
if(f[i][j][a][b])
{
int x = s[i] - '0';
f[i + 1][j + 1][(a * 10 + x) % A][b] = true;
pre[i + 1][j + 1][(a * 10 + x) % A][b] = {true, a};
f[i + 1][j][a][(b * 10 + x) % B] = true;
pre[i + 1][j][a][(b * 10 + x) % B] = {false, b};
}
int red_cnt = 50, d = 50;
for(int i = 1; i < n; i++)
{
int b = n - i;
if(f[n][i][0][0] && abs(i - b) < d) red_cnt = i, d = abs(i - b);
}
if(red_cnt == 50) puts("-1");
else
{
string ans = "";
int x = 0, y = 0;
for(int i = n; i; i--)
{
auto t = pre[i][red_cnt][x][y];
if(t.first)
{
ans.push_back('R');
red_cnt--;
x = t.second;
}
else {
ans.push_back('B');
y = t.second;
}
}
reverse(ans.begin(), ans.end());
cout << ans << endl;
}
}
return 0;
}
倒推
#include<iostream>
#include<cstring>
#include<algorithm>
#include<string>
using namespace std;
const int N = 45;
bool f[N][N][N][N];
int n, A, B;
char s[N];
string ans;
void get_pre(int n, int& cnt, int& x, int& y)
{
for(int j = 0; j <= cnt; j++)
for(int a = 0; a < A; a++)
for(int b = 0; b < B; b++)
{
int v = s[n] - '0';
if((a * 10 + v) % A == x && b == y && j + 1 == cnt && f[n][j][a][b] == f[n + 1][cnt][x][y])
{
ans.push_back('R');
cnt--;
x = a;
return;
}
else if((b * 10 + v) % B == y && a == x && j == cnt && f[n][j][a][b] == f[n + 1][cnt][x][y])
{
ans.push_back('B');
y = b;
return;
}
}
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("in.txt", "r", stdin);
#endif
int T;
cin >> T;
while(T--)
{
cin >> n >> A >> B >> s;
for(int i = 0; i <= n; i++)
for(int j = 0; j <= n; j++)
for(int a = 0; a < A; a++)
for(int b = 0; b < B; b++)
f[i][j][a][b] = false;
f[0][0][0][0] = true;
for(int i = 0; i < n; i++)
for(int j = 0; j <= i; j++)
for(int a = 0; a < A; a++)
for(int b = 0; b < B; b++)
if(f[i][j][a][b])
{
int x = s[i] - '0';
f[i + 1][j + 1][(a * 10 + x) % A][b] = true;
f[i + 1][j][a][(b * 10 + x) % B] = true;
}
int red_cnt = 50, d = 50;
for(int i = 1; i < n; i++)
{
int b = n - i;
if(f[n][i][0][0] && abs(i - b) < d) red_cnt = i, d = abs(i - b);
}
if(red_cnt == 50) puts("-1");
else
{
ans = "";
int x = 0, y = 0;
for(int i = n - 1; ~i; i--)
get_pre(i, red_cnt, x, y);
reverse(ans.begin(), ans.end());
cout << ans << endl;
}
}
return 0;
}
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