Codeforces AIM-TECH round div2 624AB 623ABC
Codeforces AIM-TECH round div2
通过数 2（一题被叉掉） 447/5602
standing change：1869-1856、
Div1的D实在没心情写，找个时间再补吧
第四题动规，第五题……好吧不知道叫什么
A：
签到不解释
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
using namespace std;
int main()
{
double s1, s2, v1, v2;
while(scanf("%lf%lf%lf%lf", &s1, &s2, &v1, &v2) != EOF){
double ans = (s2 - s1) / (v1 + v2);
printf("%.6f\n", ans);
}
return 0;
}B：
Up处特判加错WA掉
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
#define LL long long
const int MAXN = 50;
LL a[MAXN];
int main()
{
int n;
while(scanf("%d", &n) != EOF){
LL ans = 0;
for(int i = 1 ; i <= n ; i++) scanf("%I64d", &a[i]);
sort(a + 1, a + 1 + n);
LL up = 1e9 + 5;
int ok = 1;
for(int i = n ; i >= 1 ; i--){
if(up < 0){
break;
}
a[i] = min(a[i], up);
ans += a[i];
up = min(up, a[i] - 1);
// printf("i = %d, up = %I64d\n", i, up);
}
if(ok)
printf("%I64d\n", ans);
else printf("0\n");
}
return 0;
}C：
注意一下b的特殊性，就是和所有点都有连边
然后原集合最多划分成三个连通块，之后就随便写好了
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
const int MAXN = 500 + 5;
vector<int>lin[MAXN], col[5];
int g[MAXN][MAXN];
int degree[MAXN];
int color[MAXN];
int n, m;
int main()
{
while(scanf("%d%d", &n, &m) != EOF){
memset(g, 0, sizeof(g));
for(int i = 1 ; i <= 3 ; i++) col[i].clear();
for(int i = 1 ; i <= n ; i++) degree[i] = color[i] = 0, lin[i].clear();
for(int i = 0 ; i < m ; i++){
int u, v;
scanf("%d%d", &u, &v);
lin[u].push_back(v), lin[v].push_back(u);
degree[u]++, degree[v]++;
g[u][v] = g[v][u] = 1;
}
for(int i = 1 ; i <= n ; i++){
if(degree[i] == n - 1) color[i] = 2;
}
int f = 2;
int ok = 1;
for(int i = 1 ; i <= n ; i++){
if(color[i] == 0){
if(f){
int val = ((f == 2) ? 3 : 1);
color[i] = val;
col[val].push_back(i);
for(int j = 0 ; j < (int)lin[i].size() ; j++){
int v = lin[i][j];
// printf("val = %d, i = %d, v = %d\n", val, i, v);
if(color[v]) continue;
color[v] = val;
col[val].push_back(v);
}
f--;
}
else{
ok = 0;
break;
}
}
else if(color[i] != 2){
for(int j = 0 ; j < (int)lin[i].size() ; j++){
int v = lin[i][j];
if(color[v] != 2 && color[v] != color[i]){
ok = 0;
break;
}
}
}
}
if(ok){
for(int i = 1 ; i <= 3 ; i += 2){
for(int t1 = 0 ; t1 < (int)col[i].size() ; t1++){
for(int t2 = t1 + 1 ; t2 < (int)col[i].size() ; t2++){
int u = col[i][t1], v = col[i][t2];
// printf("u = %d, v = %d, i = %d\n", u, v, i);
if(g[u][v] == 0){
ok = 0;
break;
}
}
if(ok == 0) break;
}
if(ok == 0) break;
}
}
if(ok == 0) printf("No\n");
else{
printf("Yes\n");
for(int i = 1 ; i <= n ; i++){
printf("%c", 'a' + color[i] - 1);
}
printf("\n");
}
}
return 0;
}D：
这题如果分类讨论做，那你就完了
因为最后得出的集合，肯定是以某个素数k为最小因子。所以我们可以枚举这个因子。因子的来源为(a[1],a[1]-1,a[1]+1),(a[n],a[n]-1,a[n]+1)（操作和不操作）
然后dp[i][j]表示第i个，j=0表示未操作，j=1表示操作中，j=2表示操作后。
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
using namespace std;
#define LL long long
map<LL,int>mm;
const int MAXN = 1e6 + 5;
LL up;
int a[MAXN];
int n;
LL c1, c2;
LL dp[MAXN][5];
void MIN(LL &u, LL v)
{
if(u == -1 || u > v) u = v;
}
LL dfs(int v, int flag)
{
dp[0][1] = dp[0][2] = -1;
for(int i = 0 ; i < n ; i++){
// if(dp[i][0] == -1 && dp[i][1] == -1 && dp[i][2] == -1) return up + 1;
for(int j = 0 ; j < 3 ; j++) dp[i + 1][j] = -1;
// printf("i = %d, a[i] + flag = %d\n", i, a[i] + flag);
if(dp[i][0] != -1){
MIN(dp[i + 1][1], dp[i][0] + c1);
if(a[i + 1] % v == 0) MIN(dp[i + 1][0], dp[i][0]);
else if((a[i + 1] + 1) % v == 0 || ((a[i + 1] - 1) % v == 0)) MIN(dp[i + 1][0], dp[i][0] + c2);
}
if(dp[i][1] != -1){
MIN(dp[i + 1][1], dp[i][1] + c1);
if(a[i + 1] % v == 0) MIN(dp[i + 1][2], dp[i][1]);
else if((a[i + 1] + 1) % v == 0 || ((a[i + 1] - 1) % v == 0)) MIN(dp[i + 1][2], dp[i][1] + c2);
}
if(dp[i][2] != -1){
if(a[i + 1] % v == 0) MIN(dp[i + 1][2], dp[i][2]);
else if((a[i + 1] + 1) % v == 0 || ((a[i + 1] - 1) % v == 0)) MIN(dp[i + 1][2], dp[i][2] + c2);
}
}
LL temp = up + 1;
for(int i = 0 ; i < 3 ; i++) if(dp[n][i] != -1) temp = min(temp, dp[n][i]);
// if(v == 5 && flag == 1){
// printf("v = %d, flag = %d\n", v, flag);
// for(int j = 0 ; j < 3 ; j++){
// for(int i = 1 ; i <= n ; i++) printf("%I64d ", dp[i][j]);
// printf("\n");
// }
// system("pause");
// }
return temp;
}
//LL solve(int v, int flag)
//{
// LL ans = up + 1;
// for(int i = 2 ; i <= v ; i++){
// if(v % i == 0){
// ans = min(ans, dfs(i, flag));
// while(v % i == 0) v /= i;
// }
// }
// if(v > 1) ans = min(ans, dfs(v, flag));
//// printf("v = %d, flag = %d, ans = %I64d\n", v, flag, ans);
// return ans;
//}
int prime[MAXN * 10];
int cnt;
int vis[MAXN * 10];
vector<int>vc;
void init()
{
cnt = 0;
for(int i = 2 ; i < MAXN * 10 ; i++){
if(vis[i] == 0){
prime[cnt++] = i;
for(int now = i ; now < MAXN * 10 ; now += i){
vis[now] = 1;
}
}
}
}
void solve2(LL u)
{
for(int i = 0 ; i < cnt ; i++){
if(u % prime[i] == 0){
vis[i] = 1;
while(u % prime[i] == 0) u /= prime[i];
}
}
if(u > 1 && mm[u] == 0) vc.push_back(u), mm[u] = 1;
}
int main()
{
init();
while(scanf("%d%I64d%I64d", &n, &c1, &c2) != EOF){
for(int i = 1 ; i <= n ; i++) scanf("%d", &a[i]);
LL ans = c1 * (n - 1);
up = ans;
mm.clear();
vc.clear();
memset(vis, 0, sizeof(vis));
for(int i = -1 ; i <= 1 ; i++){
solve2(a[1] + i);
solve2(a[n] + i);
}
for(int i = 0 ; i < cnt ; i++){
if(vis[i]){
for(int j = -1 ; j <= 1 ; j++){
ans = min(ans, dfs(prime[i], j));
// printf("prime[i] = %d, j = %d, ans = %I64d\n", prime[i], j, ans);
}
}
}
for(int i = 0 ; i < (int)vc.size() ; i++){
for(int j = -1 ; j <= 1 ; j++){
ans = min(ans, dfs(vc[i], j));
// printf("vc[i] = %d, j = %d, ans = %I64d\n", vc[i], j, ans);
}
}
printf("%I64d\n", ans);
}
return 0;
}/*
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*/
E：
/*
这题改了三天……
刚开始以为是2SAT，好不容易调处来发现TLE
然后就用两点法做，但是由于单调性是不确定的，所以失败了
最终按照题解，二分答案判断是否正确
判断正确方法
把题目看成中间连续一段点分给x轴、其余点分给y轴
遍历所有点，对于当前点，用二分的答案来找到这个最右边点
**注意此处对于最右边点还有一个控制条件，就是它的横坐标绝对值小于等于当前遍历点，这样最后算不同轴上点距离时才不会出错
然后用预先处理好的前后缀得出此时最大最小纵坐标
找到一个合法即返回1
*/
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
#define LL long long
#define inf (200000001)
const int MAXN = 100000 + 5;
struct Point
{
LL x, y;
}d[MAXN];
bool cmp(Point u, Point v){return u.x < v.x;}
LL prefixmaxy[MAXN], prefixminy[MAXN];
LL suffixmaxy[MAXN], suffixminy[MAXN];
int n;
void init()
{
prefixmaxy[0] = suffixmaxy[n + 1] = -inf;
prefixminy[0] = suffixminy[n + 1] = inf;
for(int i = n ; i >= 1 ; i--){
suffixmaxy[i] = max(suffixmaxy[i + 1], d[i].y);
suffixminy[i] = min(suffixminy[i + 1], d[i].y);
}
for(int i = 1 ; i <= n ; i++){
prefixmaxy[i] = max(prefixmaxy[i - 1], d[i].y);
prefixminy[i] = min(prefixminy[i - 1], d[i].y);
}
}
LL sqr(LL u){return u * u;}
bool check(LL u)
{
sort(d + 1, d + 1 + n, cmp);
init();
if(sqr(d[n].x - d[1].x) <= u) return true;
if(sqr(prefixmaxy[n] - prefixminy[n]) <= u) return true;
int rear = 1;
for(int i = 1 ; i <= n ; i++){
while(i > rear) rear++;
while(rear < n && sqr(d[rear + 1].x - d[i].x) <= u && abs(d[i].x) >= abs(d[rear + 1].x)) rear++;
while(rear > i && abs(d[i].x) < abs(d[rear].x)) rear--; ///此处必须rear-1 LL miny, maxy; miny = min(prefixminy[i - 1], suffixminy[rear + 1]); maxy = max(prefixmaxy[i - 1], suffixmaxy[rear + 1]); LL tans = sqr(d[rear].x - d[i].x); tans = max(tans, sqr(maxy - miny)); tans = max(tans, max(sqr(d[rear].x), sqr(d[i].x)) + max(sqr(maxy), sqr(miny)) ); if(tans <= u) return true; } for(int i = 1 ; i <= n ; i++) d[i].x = -d[i].x; ///更改大小关系
sort(d + 1, d + 1 + n, cmp);
init();
rear = 1;
for(int i = 1 ; i <= n ; i++){
while(i > rear) rear++;
while(rear < n && sqr(d[rear + 1].x - d[i].x) <= u && abs(d[i].x) >= abs(d[rear + 1].x)) rear++;
while(rear > i && abs(d[i].x) < abs(d[rear].x)) rear--;
LL miny, maxy;
miny = min(prefixminy[i - 1], suffixminy[rear + 1]);
maxy = max(prefixmaxy[i - 1], suffixmaxy[rear + 1]);
LL tans = sqr(d[rear].x - d[i].x);
tans = max(tans, sqr(maxy - miny));
tans = max(tans, max(sqr(d[rear].x), sqr(d[i].x)) + max(sqr(maxy), sqr(miny)) );
if(tans <= u) return true;
}
return false;
}
int main()
{
while(scanf("%d", &n) != EOF){
for(int i = 1 ; i <= n ; i++) scanf("%I64d%I64d", &d[i].x, &d[i].y);
sort(d + 1, d + 1 + n, cmp);
if(n == 1){
printf("0\n");
continue;
}
LL le = 0, re = (LL)inf * inf;
while(le < re - 1){
LL mid = (le + re) >> 1;
if(check(mid)) re = mid;
else le = mid;
}
LL ans;
if(check(le)) ans = le;
else ans = re;
printf("%I64d\n", ans);
}
return 0;
}