B Buffoon
Determining a first maximum number is not, check title.
H Hour for a Run
Output \ (n * m \) a \ (10 \% \) to \ (90 \% \) , sign in question, and be careful not to use floating-point ceil, have accuracy problems.
M Brazilian Marathon Popcorn
Reading a bit difficult problem, is to give the number n, m individuals, each sub-section a continuous, each number can only be given to a person, the number per second up Save k, Q minimum time.
Two points on the boundary of each segment is determined to greed.
#include <bits/stdc++.h>
using namespace std;
const int N=1e5+50;
int n,m,k,a[N];
bool check(int x){
int c=1;
int tmp=0;
for(int i=1;i<=n;i++){
if(a[i]>x){
return false;
}
if(tmp+a[i]<=x){
tmp+=a[i];
}else{
c++;
tmp=a[i];
}
}
return c<=m;
}
int main(){
scanf("%d%d%d",&n,&m,&k);
int sum=0;
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
sum+=a[i];
}
int l=1,r=sum;
int ans=0;
while(l<=r){
int mid=(l+r)/2;
if(check(mid)){
r=mid-1;
ans=mid;
}else{
l=mid+1;
}
}
printf("%d\n",(ans-1)/k+1);
return 0;
}
D Denouncing Mafia
Corresponding to a given question are intended tree, to find that most point k strand coverage.
Consider greedy, be sure to take every one of the longest chain, then took after this chain, the tree becomes a forest, next time out longest chain from these trees, then add some trees
dfs over a pretreatment corresponding to each node in the subtree root node and forest longest chain after the chain has been omitted longest, with a priority queue can take greedy.
#include <bits/stdc++.h>
using namespace std;
const int N=1e5+55;
vector<int> g[N];
int ld[N];
vector<int> ls[N];
int n,k,f;
struct node{
int u,w;
bool operator<(const node& rhs)const{
return w<rhs.w;
}
};
void dfs(int u){
int siz=g[u].size();
if(!siz){
ld[u]=1;
return;
}
int mx=0;
int k=0;
for(int i=0;i<siz;i++){
int v=g[u][i];
dfs(v);
if(ld[v]>mx){
mx=ld[v];
k=v;
}
}
ld[u]=mx+1;
for(int i=0;i<siz;i++){
int v=g[u][i];
if(v!=k){
ls[u].push_back(v);
}
}
int vs=ls[k].size();
for(int j=0;j<vs;j++){
ls[u].push_back(ls[k][j]);
}
}
int main(){
//freopen("in.txt","r",stdin);
scanf("%d%d",&n,&k);
for(int i=2;i<=n;i++){
scanf("%d",&f);
g[f].push_back(i);
}
dfs(1);
priority_queue<node> pq;
pq.push({1,ld[1]});
int ans=0;
while(!pq.empty()){
auto t=pq.top();
int u=t.u;
int w=t.w;
pq.pop();
ans+=w;
k--;
if(!k){
break;
}
int siz=ls[u].size();
for(int i=0;i<siz;i++){
pq.push({ls[u][i],ld[ls[u][i]]});
}
}
printf("%d\n",ans);
return 0;
}
L Less Coin Tosses
Question is intended essentially all 01 strings of length n, the same number of 1 and 0 can be matched to find the string does not match the number 01.
Enumeration consider 0 (or 1) the number of such strings have \ (C_n ^ i \) a, then the number of the strings 01 is obviously the same, if an even number, can all match, if it is an odd number, the answer plus 1.
So essentially the request \ (\ sum_ {I}. 1 = n-C_n ^ ^ I \% 2 \) , play table to find the law, we have found a common routine recursive doubling.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n;
ll pw[115],f[115];
int cnt;
void init(){
pw[0]=1;
f[0]=pw[0]-1;
for(int i=1;i<=105;i++){
pw[i]=pw[i-1]*2;
f[i]=pw[i]-1;
if(pw[i]>(ll)1e18){
cnt=i;
break;
}
}
}
ll solve(int k,ll n){
if(k==1){
return 2ll;
}
ll mid=pw[k-1]/2;
if(n<=mid){
return solve(k-1,n);
}else{
return 2ll*solve(k-1,n-mid);
}
}
int main(){
init();
scanf("%lld",&n);
int k=lower_bound(f+1,f+1+cnt,n)-f;
ll ans=solve(k,n-f[k-1]);
printf("%lld\n",ans);
return 0;
}
G Getting Confidence
Given \ (n * n \) matrix, each column of a selected row number and can not be the same, so that the product of the maximum.
- Did a similar routine title, will find a number on the number, then the product is the biggest and largest, network flow and map building, limiting their number per row, running a minimum cost maximum flow.
#include <bits/stdc++.h>
using namespace std;
const int N=1e5+50;
const int INF=0x3f3f3f3f;
const double DINF=1.0*1e18;
const double eps=1e-8;
struct Edge{
int u,v,w;
double c;
int next;
}e[N];
int ns,n,mt[105][105];
int cnt,head[N];
void init(){
cnt=0;
memset(head,-1,sizeof(head));
}
void add(int u,int v,int w,double c){
e[cnt]=Edge{u,v,w,c,head[u]};
head[u]=cnt++;
e[cnt]=Edge{v,u,0,-c,head[v]};
head[v]=cnt++;
}
double d[N];
int inq[N],s,t,p[N],a[N];
bool bf(int &flow,double &cost){
for(int i=0;i<ns;i++){
d[i]=DINF;
inq[i]=0;
}
d[s]=0;
p[s]=0;
a[s]=INF;
queue<int> q;
q.push(s);
inq[s]=1;
while(!q.empty()){
int u=q.front();
q.pop();
inq[u]=false;
for(int i=head[u];i!=-1;i=e[i].next){
int v=e[i].v;
int w=e[i].w;
double c=e[i].c;
if(w>0 && d[v]>d[u]+c){
d[v]=d[u]+c;
p[v]=i;
a[v]=min(a[u],w);
if(!inq[v]){
q.push(v);
inq[v]=1;
}
}
}
}
if(fabs(d[t]-DINF)<eps){
return 0;
}
flow+=a[t];
cost+=d[t]*a[t];
for(int u=t;u!=s;u=e[p[u]].u){
e[p[u]].w-=a[t];
e[p[u]^1].w+=a[t];
}
return 1;
}
void mcmf(int &flow,double &cost){
flow=0;
cost=0.0;
while(bf(flow,cost));
}
int main(){
//freopen("in.txt","r",stdin);
scanf("%d",&n);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
scanf("%d",&mt[i][j]);
}
}
ns=(n+2)*n+2;
s=0,t=ns-1;
init();
for(int i=1;i<=n;i++){
add(s,n*n+i,1,0.0);
}
for(int i=1;i<=n;i++){
add(n*n+n+i,t,1,0.0);
}
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
add(n*n+i,(j-1)*n+i,1,0.0);
add((j-1)*n+i,n*n+n+j,1,-log10(mt[j][i]));
}
}
int flow=0;
double cost=0;
mcmf(flow,cost);
for(int i=n*n+1;i<=n*n+n;i++){
for(int j=head[i];j!=-1;j=e[j].next){
if(e[j].w==0){
printf("%d",(e[j].v-1)/n+1);
if(i==n*n+n){
printf("\n");
}else{
printf(" ");
}
break;
}
}
}
return 0;
}J Jar of Water Game
Meaning of the questions unclear simulation title ... Fortunately, his teammates what cards are played
n individuals, each person has four cards, initially designated the k individuals have a wild card from him and started every election appears a minimum number of cards to the next person, and if there is not a wild card just to get, you must the wild card to the next person, if there is more than the number of cards appear as small, giving little value.
Victory is defined as the state of the hands of the same brand 4, you can not have a wild card , if the number of people the state of victory, that is the value of the same brand of small victory.
#include <bits/stdc++.h>
using namespace std;
int idx(char c){
if(c=='A'){
return 1;
}else if(c>='2' && c<='9'){
return c-'0';
}else if(c=='D'){
return 10;
}else if(c=='Q'){
return 11;
}else if(c=='J'){
return 12;
}else if(c=='K'){
return 13;
}else if(c=='X'){
return 14;
}
}
int n,k;
struct node{
int cnt[15];
int tm;
bool has;
}a[1005];
char s[10];
int win(){
int ans=0;
int res=0x3f3f3f3f;
for(int x=1;x<=n;x++){
int mx=0;
int mk=0;
for(int i=1;i<=13;i++){
mx=max(mx,a[x].cnt[i]);
if(a[x].cnt[i]>mx){
mx=a[x].cnt[i];
mk=i;
}
}
if(mx==4 && !a[x].has){
if(mk<res){
res=mk;
ans=x;
}
}
}
return ans;
}
int get(int x){
if(a[x].has && a[x].tm==1){
return 14;
}else{
if(a[x].has){
a[x].tm=1;
}
int mn=0x3f3f3f3f;
for(int i=1;i<=13;i++){
if(!a[x].cnt[i]){
#include <bits/stdc++.h>
using namespace std;
int idx(char c){
if(c=='A'){
return 1;
}else if(c>='2' && c<='9'){
return c-'0';
}else if(c=='D'){
return 10;
}else if(c=='Q'){
return 11;
}else if(c=='J'){
return 12;
}else if(c=='K'){
return 13;
}else if(c=='X'){
return 14;
}
}
int n,k;
struct node{
int cnt[15];
int tm;
bool has;
}a[1005];
char s[10];
int win(){
int ans=0;
int res=0x3f3f3f3f;
for(int x=1;x<=n;x++){
int mx=0;
int mk=0;
for(int i=1;i<=13;i++){
mx=max(mx,a[x].cnt[i]);
if(a[x].cnt[i]>mx){
mx=a[x].cnt[i];
mk=i;
}
}
if(mx==4 && !a[x].has){
if(mk<res){
res=mk;
ans=x;
}
}
}
return ans;
}
int get(int x){
if(a[x].has && a[x].tm==1){
return 14;
}else{
if(a[x].has){
a[x].tm=1;
}
int mn=0x3f3f3f3f;
for(int i=1;i<=13;i++){
if(!a[x].cnt[i]){
continue;
}
mn=min(mn,a[x].cnt[i]);
}
for(int i=1;i<=13;i++){
if(mn==a[x].cnt[i]){
return i;
}
}
}
}
int main(){
//freopen("in.txt","r",stdin);
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++){
scanf("%s",s);
for(int j=0;j<4;j++){
int x=idx(s[j]);
a[i].cnt[x]++;
}
a[i].has=false;
a[i].tm=0;
}
a[k].has=true;
int cur=k;
while(true){
if(int w=win()){
printf("%d\n",w);
return 0;
}
int nex=get(cur);
int nt=cur%n+1;
a[cur].cnt[nex]--;
a[nt].cnt[nex]++;
if(nex==14){
a[cur].has=false;
a[cur].tm=0;
a[nt].has=true;
a[nt].tm=0;
}
cur=nt;
}
return 0;
}
continue;
}
mn=min(mn,a[x].cnt[i]);
}
for(int i=1;i<=13;i++){
if(mn==a[x].cnt[i]){
return i;
}
}
}
}
int main(){
//freopen("in.txt","r",stdin);
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++){
scanf("%s",s);
for(int j=0;j<4;j++){
int x=idx(s[j]);
a[i].cnt[x]++;
}
a[i].has=false;
a[i].tm=0;
}
a[k].has=true;
int cur=k;
while(true){
if(int w=win()){
printf("%d\n",w);
return 0;
}
int nex=get(cur);
int nt=cur%n+1;
a[cur].cnt[nex]--;
a[nt].cnt[nex]++;
if(nex==14){
a[cur].has=false;
a[cur].tm=0;
a[nt].has=true;
a[nt].tm=0;
}
cur=nt;
}
return 0;
}