Educational Codeforces Round 46
A. Codehorses T-shirts
题目大意：给出n个上一年的衣服码数和n个今年的衣服码数，问最少修改多少次字母把上一年的衣服码数变成今年的衣服码数
题解：开一个两个索引的map，一个索引是长度，另一个索引是最后一个字母，模拟一下就好了
#include
#define FOR(i, a, b) for(int i=(a); i<(b); i++)
#define REP(i, a, b) for(int i=(a); i<=(b); i++)
#define DOWN(i, a, b) for(int i=(a); i>=(b); i--)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair pll;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1000000009;
const int N = 1e3 + 10;
int n;
string s;
map<int, map<char, int>> M, M2;
int len;
int ans;
int main() {
cin>>n;
REP(i, 1, n) {
cin>>s;
len=s.length();
if (len==1) M[len][s[0]]++;
else M[len][s.back()]++;
}
REP(i, 1, n) {
cin>>s;
len=s.length();
if (len==1) M2[len][s[0]]++;
else M2[len][s.back()]++;
}
if (M[1]['M']>M2[1]['M']) ans+=(M[1]['M']-M2[1]['M']);
if (M[1]['S']>M2[1]['S']) ans+=(M[1]['S']-M2[1]['S']);
if (M[1]['L']>M2[1]['L']) ans+=(M[1]['L']-M2[1]['L']);
ans+=abs(M[2]['S']-M2[2]['S']);
ans+=abs(M[3]['S']-M2[3]['S']);
ans+=abs(M[4]['S']-M2[4]['S']);
cout<//cout<<1.*clock()/CLOCKS_PER_SEC<<"ms"<<"\n";
return 0;
}B.Light It Up
题目大意：给出n个时间点，每个时间点灯都会改变开关状态，问插入一个时间点后，亮灯时间最长为多长
题解：先把0，M也插入进去，可以证明，插入点必为序号为偶数的点+1或-1，先预处理出一开始的亮灯时间，每次枚举插入时间点后翻转区间就好了
#include
#define FOR(i, a, b) for(int i=(a); i<(b); i++)
#define REP(i, a, b) for(int i=(a); i<=(b); i++)
#define DOWN(i, a, b) for(int i=(a); i>=(b); i--)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair pll;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1000000009;
const int N = 1e5 + 10;
int n, M;
int a[N];
set<int> S;
int sum[N];
int t, ans;
int len;
int L, R;
int main() {
cin>>n>>M;
a[1]=0;
a[n+2]=M;
REP(i, 2, n+1) cin>>a[i];
n=n+2;
S.insert(0);
S.insert(M);
REP(i, 1, n) {
S.insert(a[i]);
if (i%2==1) sum[i]=sum[i-1];
else sum[i]=sum[i-2]+a[i]-a[i-1];
}
ans=sum[n];
REP(i, 1, n) {
t=a[i]-1;
if (t<0||t>M) continue;
if (S.count(t)) continue;
if (i%2==1) continue;
L=sum[i]-1;
R=sum[n]-sum[i]+1;
len=M-t;
ans=max(ans, L+len-R);
}
REP(i, 1, n) {
t=a[i]+1;
if (t<0||t>M) continue;
if (S.count(t)) continue;
if (i%2==1) continue;
L=sum[i];
R=sum[n]-sum[i];
len=M-t;
ans=max(ans, L+len-R);
}
cout<//cout<<1.*clock()/CLOCKS_PER_SEC<<"ms"<<"\n";
return 0;
}C. Covered Points Count
题目大意：给出n个线段，问覆盖i次的点有多少个
题解：扫描线经典题目
#include
#define FOR(i, a, b) for(int i=(a); i<(b); i++)
#define REP(i, a, b) for(int i=(a); i<=(b); i++)
#define DOWN(i, a, b) for(int i=(a); i>=(b); i--)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair pll;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1000000009;
const int N = 4e6 + 10;
int n;
ll l, r;
int cnt;
int k;
ll len;
ll ans[N];
struct node {
ll x;
int p;
bool operator <(const node& rhs) const {
return xint main() {
cin>>n;
REP(i, 1, n) {
cin>>l>>r;
a[cnt].p=1;
a[cnt++].x=l;
a[cnt].p=-1;
a[cnt++].x=r+1;
}
sort(a, a+cnt);
FOR(i, 0, cnt) {
if (i>0) {
len=a[i].x-1-a[i-1].x+1;
ans[k]+=len;
}
k+=a[i].p;
}
REP(i, 1, n) {
cout<if (i==n) cout<else cout<<' ';
}
//cout<<1.*clock()/CLOCKS_PER_SEC<<"ms"<<"\n";
return 0;
} D. Yet Another Problem On a Subsequence
题目大意：一个数组为好数组当且仅当这个数组有k个元素且首元素为k-1，一个序列为好序列当且仅当这个序列可以分为几个好数组
题解：定义dp[i]为以a[i]开头的好序列个数，容易发现dp[i]的递推关系
#include
#define FOR(i, a, b) for(int i=(a); i<(b); i++)
#define REP(i, a, b) for(int i=(a); i<=(b); i++)
#define DOWN(i, a, b) for(int i=(a); i>=(b); i--)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair pll;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
const int N = 1e4 + 10;
ll f[N], inv[N], a[N], dp[N];
int n;
ll ans;
ll C(ll n, ll m) {
if (nreturn 0;
return f[n]*inv[n-m]%mod*inv[m]%mod;
}
ll Pow(ll a, ll b) {
ll ret=1;
while(b) {
if (b&1) ret=ret*a%mod;
a=a*a%mod;
b>>=1;
}
return ret;
}
void init() {
f[0]=1;
REP(i, 1, 2000) {
f[i]=f[i-1]*i%mod;
}
REP(i, 0, 2000) {
inv[i]=Pow(f[i], mod-2);
}
}
int main() {
init();
cin>>n;
REP(i, 1, n) cin>>a[i];
DOWN(i, n, 1) {
if (a[i]<=0) continue;
dp[i]=C(n-i, a[i]);
REP(j, i+1, n) {
if (a[j]<=0) continue;
dp[i]=(dp[i]+(dp[j]*C(j-i-1, a[i]))%mod)%mod;
}
}
REP(i, 1, n) ans=(ans+dp[i])%mod;
cout<//cout<<1.*clock()/CLOCKS_PER_SEC<<"ms"<<"\n";
return 0;
} 教练，我想打ACM