The 10th Shandong Provincial Collegiate Programming Contest（8/13）
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[The 10th Shandong Provincial Collegiate Programming Contest]
(A.Calandar)
签到
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
char day[5][20] = {{"Monday"},{"Tuesday"},{"Wednesday"},{"Thursday"},{"Friday"}};
using LL = int_fast64_t;
int T;
void solve(){
LL now = 0, y, m, d, nxt = 0;
int cd;
char s[20];
scanf("%I64d %I64d %I64d %s",&y,&m,&d,s);
if(s[0]=='M') cd = 0;
else if(s[0]=='T' && s[1]=='u') cd = 1;
else if(s[0]=='W') cd = 2;
else if(s[0]=='T' && s[1]=='h') cd = 3;
else cd = 4;
now = 360 * y + 30 * m + d;
scanf("%I64d %I64d %I64d",&y,&m,&d);
nxt = 360 * y + 30 * m + d;
LL delta = nxt - now;
cd = ((cd+delta)%5+5)%5;
printf("%s
",day[cd]);
}
int main(){
for(scanf("%d",&T); T; T--) solve();
return 0;
}(B.Flipping Game)
DP
(f[i][j])表示当前第(i)轮按开关，和最终状态相反的有(j)个的方案数，初始有(num)个灯和最终状态不同的话，显然(f[0][num]=1)
现在考虑状态转移，(kk)表示上一轮与最终状态相反的个数，(x)表示有(x)个从和最终状态相反的变成和最终状态相同的，(y)表示从和最终状态相同的变成和最终状态不同的，则(f[i][j]=sum_{valid kk}f[i-1][kk]·C(kk,x)·C(n-kk,y))
其中需要满足四个约束条件：
1.(0 le x le kk)
2.(0 le y le n-kk)
3.(x + y = m)
4.(kk-x+y=j)
(3)和(4)联立得到(x=(kk+m-j)/2)
然后就可以写了
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 111;
using LL = int_fast64_t;
const LL MOD = 998244353;
int T,n,m,k;
char s1[MAXN],s2[MAXN];
LL fact[MAXN],invfact[MAXN],f[2][MAXN];
LL qpow(LL a, LL b){
LL res = 1;
while(b){
if(b&1) res = res * a % MOD;
b >>= 1;
a = a * a % MOD;
}
return res;
}
LL C(int A, int B){ return fact[A] * invfact[B] % MOD * invfact[A-B] % MOD; }
void solve(){
scanf("%d %d %d %s %s",&n,&k,&m,s1,s2);
int num = 0;
for(int i = 0; i < n; i++) num += (s1[i]!=s2[i]?1:0);
memset(f,0,sizeof(f));
int tag = 0;
f[tag][num] = 1;
for(int i = 1; i <= k; i++){
tag ^= 1;
memset(f[tag],0,sizeof(f[tag]));
for(int j = 0; j <= n; j++){
for(int kk = max(0,j-m); kk <= min(n,j+m); kk++){
if((kk+m-j)&1) continue;
int x = ((kk+m-j)>>1);
if(x<0||x>kk||m-x<0||m-x>n-kk) continue;
f[tag][j] = (f[tag][j] + f[tag^1][kk] * C(kk,x) % MOD * C(n-kk,m-x)) % MOD;
}
}
}
printf("%lld
",f[tag][0]);
}
int main(){
fact[0] = 1;
for(int i = 1; i < MAXN; i++) fact[i] = fact[i-1] * i % MOD;
for(int i = 0; i < MAXN; i++) invfact[i] = qpow(fact[i],MOD-2);
for(scanf("%d",&T); T; T--) solve();
return 0;
}(C.Wandering Robot)
随便写
#include<bits/stdc++.h>
#pragma GCC optimize("O3")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
using namespace std;
#define IOS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
const int N = 1e5+10;
char a[N];
int32_t main(){
IOS;
int _;cin>>_;
while(_--){
int n,k; cin>>n>>k;
cin>>a;
int ans=0,x=0,y=0;
for(int i=0;i<n;i++){
if(a[i]=='R')x++;
if(a[i]=='L')x--;
if(a[i]=='U')y++;
if(a[i]=='D')y--;
ans=max(ans, abs(x)+abs(y));
}
if(k==1){
cout<<ans<<endl;
continue;
}
int xx=x*(k-1), yy=y*(k-1);
for(int i=0;i<n;i++){
if(a[i]=='R')xx++;
if(a[i]=='L')xx--;
if(a[i]=='U')yy++;
if(a[i]=='D')yy--;
ans=max(ans, abs(xx)+abs(yy));
}
cout<<ans<<endl;
}
return 0;
}(D.Game on a Graph)
最终状态是一棵树，只和边的数量有关，最多可以删掉(e-(v-1))条边
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e5+7;
int T,n,v,e;
char s[MAXN];
void solve(){
scanf("%d %s %d %d",&n,s,&v,&e);
for(int i = 1; i <= e; i++) scanf("%d %d",&n,&n);
n = strlen(s);
putchar(s[(e-(v-1))%n]=='1'?'2':'1'); puts("");
}
int main(){
for(scanf("%d",&T); T; T--) solve();
return 0;
}(E.BaoBao Loves Reading)
(F.Stones in the Bucket)
贪心，先变成(n)的倍数然后把低于平均数的补齐即可
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
using LL = int_fast64_t;
const int MAXN = 1e5+7;
int T,n;
LL A[MAXN];
void solve(){
LL res = 0;
cin >> n;
for(int i = 1; i <= n; i++) cin >> A[i];
LL tot = accumulate(A+1,A+1+n,0LL);
LL del = tot % n, ave = tot / n;
res = del;
for(int i = 1; i <= n; i++){
if(A[i]>ave&&del){
LL d = min(del,A[i]-ave);
del -= d;
A[i] -= d;
}
if(!del) break;
}
for(int i = 1; i <= n; i++) if(A[i]<ave) res += ave - A[i];
cout << res << endl;
}
int main(){
____();
for(cin >> T; T; T--) solve();
return 0;
}(G.Heap)
(H.Tokens on the Segments)
优先考虑左端点靠前的，每当遍历到某(x)值时，把左端点为(x)的线段的右端点加入到优先队列中，把已经超过右边界的线段删去，然后取右端点最靠左的放token
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e5+7;
int T,n;
pair<int,int> seg[MAXN];
void solve(){
cin >> n;
for(int i = 1; i <= n; i++) cin >> seg[i].first >> seg[i].second;
sort(seg+1,seg+1+n);
int tot = 0, now = 1;
priority_queue<int,vector<int>,greater<int>> que;
for(int cur = 1; now<=n||!que.empty(); ){
while(now<=n&&seg[now].first==cur){
que.push(seg[now].second);
now++;
}
while(!que.empty()&&que.top()<cur) que.pop();
if(!que.empty()){
tot++;
que.pop();
}
if(!que.empty()) cur++;
else{
if(now>n) break;
else cur = seg[now].first;
}
}
cout << tot << endl;
}
int main(){
____();
for(cin >> T; T; T--) solve();
return 0;
}(I.Connected Intervals)
(J.Triangle City)
(K.Happy Equation)
(L.Median)
建大的向小的连边，同时建反图，如果图中存在环就直接全部输出0.
找每个点在拓扑排序序列中能够达到的最左端和最右端，如果((n+1)/2)在区间内的话，说明可以作为中位数，用bitset维护最少必经集合
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 111;
int T,n,m,deg[MAXN],rdeg[MAXN];
vector<int> G[MAXN],rG[MAXN];
bitset<MAXN> bst[MAXN],rbst[MAXN];
int toposort(){
int tot = 0;
queue<int> que;
for(int i = 1; i <= n; i++) if(!deg[i]) que.push(i);
while(!que.empty()){
int u = que.front();
que.pop();
tot++;
for(int v : G[u]){
if(--deg[v]==0) que.push(v);
bst[v] |= bst[u];
}
}
return tot;
}
void rtoposort(){
queue<int> que;
for(int i = 1; i <= n; i++) if(!rdeg[i]) que.push(i);
while(!que.empty()){
int u = que.front();
que.pop();
for(int v : rG[u]){
if(--rdeg[v]==0) que.push(v);
rbst[v] |= rbst[u];
}
}
}
void init(){
for(int i = 1; i <= n; i++){
G[i].clear();
rG[i].clear();
rdeg[i] = 0;
deg[i] = 0;
bst[i].reset();
bst[i].set(i);
rbst[i].reset();
rbst[i].set(i);
}
}
void solve(){
scanf("%d %d",&n,&m);
init();
for(int i = 1; i <= m; i++){
int u, v;
scanf("%d %d",&u,&v);
G[u].emplace_back(v);
rG[v].emplace_back(u);
deg[v]++;
rdeg[u]++;
}
int tot = toposort();
if(tot!=n){
for(int i = 0; i < n; i++) putchar('0');
puts("");
return;
}
rtoposort();
for(int i = 1; i <= n; i++){
int l = bst[i].count();
int r = n-rbst[i].count()+1;
if(l<=(n+1)/2 && (n+1)/2<=r) putchar('1');
else putchar('0');
}
puts("");
}
int main(){
for(scanf("%d",&T); T; T--) solve();
return 0;
}(M. Sekiro)
签到
include<bits/stdc++.h>
using namespace std;
define IOS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
const int N = 1e5+10;
int main(){
IOS;
int t; cin>>t;
while(t--){
int n,k; cin>>n>>k;
if(!n){
cout<<0<<endl;
continue;
}
while(k--){
if(n&1)n++;
n>>=1;
if(n==1)break;
}
cout<<n<<endl;
}
return 0;
}
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