A
- Meaning of the questions: garbage
- Ideas: Analog
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <stdlib.h>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define ll long long
#define FOR(i,l,r) for(int i = l ; i <= r ;++i )
#define inf 1<<30
#define EPS (1e-9)
#define ALL(T) T.begin(),T.end()
#define lson(i) i<<1
#define rson(i) (i<<1|1)
using namespace std;
typedef pair<int,int> pii;
int cnt[10];
char buf[3010];
char ma[1010];
int main(){
int t;
cin >> t;
FOR(cas,1,t){
scanf("%s%s",buf,ma);
cnt[1] = cnt[2] = cnt[3] = 0;
int len = strlen(buf);
for(int i=0;i<len;++i){
switch (ma[buf[i]-'a'])
{
case 'h':
cnt[1]++;
break;
case 'w':
cnt[2]++;
break;
case 'd':
cnt[3]++;
break;
default:
break;
}
}
printf("Case #%d: ",cas);
if(cnt[1]>=(len/4.0)){
printf("Harmful\n");
}else if(cnt[1]<=(len/10.0)){
printf("Recyclable\n");
}else if(cnt[3]>=cnt[2]*2){
printf("Dry\n");
}else{
printf("Wet\n");
}
}
return 0;
}To double divisorOr will wa QAQ
B
- Question is intended: to convert the binary address into ipv6 shortest lexicographically smallest standard format (requires reduction 0)
- Ideas: analog, the first converted into hex, then remove the front 0, 0 last contraction through all the cases, get answers.
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <stdlib.h>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define ll long long
#define FOR(i,l,r) for(int i = l ; i <= r ;++i )
#define inf 1<<30
#define EPS (1e-9)
#define ALL(T) T.begin(),T.end()
#define lson(i) i<<1
#define rson(i) (i<<1|1)
using namespace std;
typedef pair<int,int> pii;
char buf[1029];
char ox[110];
char ox2[110];
char ansStr[110];
void toOx(){ // 变成16进制
int idx = 0;
int val = 0;
int oidx = 0;
while(idx<128){
val= (val<<1)+(buf[idx++]-'0');
val= (val<<1)+(buf[idx++]-'0');
val= (val<<1)+(buf[idx++]-'0');
val= (val<<1)+(buf[idx++]-'0');
if(val<10){
ox[oidx] = val+'0';
}else{
ox[oidx] = val-10+'a';
}
oidx++;
val = 0;
}
ox[oidx] = 0;
oidx = 0;
FOR(i,0,31){
ox2[oidx++] = ox[i];
}
ox2[oidx] = 0;
idx = oidx;
oidx = 0;
FOR(i,0,idx-1){
if(ox2[i]=='0'){
if(ox2[i+1]=='0'){
if(ox2[i+2]=='0'){
ox[oidx++] = ox2[i+3];
ox[oidx++] = ':';
i+=3;
}else{
ox[oidx++] = ox2[i+2];
ox[oidx++] = ox2[i+3];
ox[oidx++] = ':';
i+=3;
}
}else{
ox[oidx++] = ox2[i+1];
ox[oidx++] = ox2[i+2];
ox[oidx++] = ox2[i+3];
ox[oidx++] = ':';
i+=3;
}
}else{
ox[oidx++] = ox2[i];
ox[oidx++] = ox2[i+1];
ox[oidx++] = ox2[i+2];
ox[oidx++] = ox2[i+3];
ox[oidx++] = ':';
i+=3;
}
}
ox[oidx-1] = 0;
// printf("%s\n%s\n",ox2,ox);
}
int cmp(string s1,string s2){
if(s1.length() == s2.length()){
return s1<s2;
}
return s1.length() <s2.length();
}
vector<string> as;
void ya0(){ // 压缩0
vector<pii > pos;
int len = strlen(ox);
pii lin;
FOR(i,0,len-1){
if(ox[i]=='0'&&ox[i+1]==':'&&(i-1<0 ||ox[i-1]==':')){
lin.second = i;
lin.first = 0;
while(i+2<len&&ox[i+2]=='0'){
i+=2;
lin.first++;
}
pos.push_back(lin);
}
}
int idx= 0;
sort(ALL(pos));
if(pos.size()==0 || pos[pos.size()-1].first==0){
FOR(i,0,len){
ansStr[i] = ox[i];
}
as.push_back(ansStr);
return ;
}
int tmp = pos[pos.size()-1].first;
for(auto lin:pos){
if(lin.first!=tmp) continue;
idx = 0;
FOR(i,0,len-1){
if(i!=lin.second){
ansStr[idx++] = ox[i];
}else{
if(ansStr[idx-1]!=':')
ansStr[idx++] = ':';
ansStr[idx++] = ':';
i += lin.first*2+1;
}
}
ansStr[idx] = 0;
// string tstr = ansStr;
as.push_back(ansStr);
}
sort(ALL(as),cmp);
}
int main(){
int t;
scanf("%d",&t);
FOR(cas,1,t){
as.clear();
scanf("%s",buf);
toOx();
ya0();
printf("Case #%d: ",cas);
cout << as[0] << '\n';
}
return 0;
}Zero Zero pressure and intermediate pressure end to end, the string length is not the same direct sort through all circumstances it again
D
- Meaning of the questions: There are boxes k, n two items, each item has a volume \ (a_i \) and put up items to go under a box for each box to be put into not seek to set aside the smallest of all. box capacity items
- Thinking: From the lower bound \ (\ frac {\ sum { a_i}} {n} \) begins the enumeration of the size of the box, multiset of upper_bound, (greater than the first one) to simulate the process of discharge
#include<bits/stdc++.h>
#define ll long long
#define FOR(i,l,r) for(int i = l ; i <= r ;++i )
#define inf 1<<30
#define EPS (1e-9)
#define ALL(T) T.begin(),T.end()
#define lson(i) i<<1
#define rson(i) (i<<1|1)
using namespace std;
const int maxn(1e4+5);
int a[maxn],r,l;
int n,k;
void input(){
memset(a,0,sizeof(a));
scanf("%d%d",&n,&k);
r = 0;
FOR(i,1,n){
scanf("%d",&l);
r += l;
a[l]++;
}
r/=k;
}
multiset<int> s;
int check(int mid){
s.clear();
for(int i=1;i<=1000;++i){
for(int j=1;j<=a[i];++j){
s.insert(i);
}
}
for(int i=1;i<=k;++i){
int y = mid;
while(y){
auto iter = s.upper_bound(y);
if(iter != s.begin()){
--iter;
y -=(*iter);
s.erase(iter);
}else{
break;
}
}
}
return s.size()==0;
}
int main(){
int t;
scanf("%d",&t);
FOR(cas,1,t){
input();
while(!check(r)){
r++;
}
printf("Case #%d: %d\n",cas,r);
}
return 0;
}This question will readily occur to half, thus thought \ (O (n ^ 2) \) a \ (Check () \) + \ (O (log (T)) \) of the two points, but this value is not continuous the question is not satisfied to fill half of the property ... can be half the interval through every period of enumeration.
there was a very similar topic U33405 New York
G
- The meaning of problems: given letter n \ (A to J \) character string, if there is a demand \ (0 to 9 \) arranged all strings can be mapped to a valid date and is Friday.
- Thinking: all pre-arranged first, and then enumerated by Zeller's congruence is determined whether a legitimate
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <stdlib.h>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define ll long long
#define FOR(i,l,r) for(int i = l ; i <= r ;++i )
#define inf 1<<30
#define EPS (1e-9)
#define ALL(T) T.begin(),T.end()
#define lson(i) i<<1
#define rson(i) (i<<1|1)
using namespace std;
typedef pair<int,int> pii;
const int maxn = 1e6+10;
int mon[15] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int whatDay(int d,int m,int y){
int ans;
if(m==1||m==2)
m+=12,y--;
ans = (d+2*m+3*(m+1)/5 + y + y/4 - y/100 + y/400 + 1)%7;
return ans == 5;
}
int check(string op,int* a){
int y = a[op[0]-'A']*1000 + a[op[1]-'A']*100 + a[op[2]-'A']*10 + a[op[3]-'A'];
int m = a[op[5]-'A']*10 + a[op[6]-'A'];
int d = a[op[8]-'A']*10 + a[op[9]-'A'];
if(y%4==0 && y%100 || y%400==0) mon[2] = 29;
else mon[2] = 28;
if(y < 1600 || y>9999 || m<1 || m>12 || d<1 || d>mon[m]) return 0;
return whatDay(d,m,y);
}
int p[3628800][10];
void init(){
int a[10];
for(int i=0;i<10;++i)
a[i] = i;
int cnt = 0;
do{
memcpy(p[cnt++],a,sizeof(a));
}while(next_permutation(a,a+10));
}
string buf[maxn];
vector<int> tr;
int main(){
int t,n;
init();
cin >> t;
FOR(cas,1,t){
cin >> n;
FOR(i,1,n){
cin >> buf[i];
}
sort(buf+1,buf+n);
n = unique(buf+1,buf+n+1)-buf-1;
random_shuffle(buf+1,buf+n+1);
printf("Case #%d: ",cas);
int ok = 0;
FOR(i,0,3628800-1){
ok = 1;
FOR(j,1,n){
if(!check(buf[j],p[i])){
ok = 0;
break;
}
}
if(ok){
// tr.push_back(i);
FOR(j,0,9){
printf("%d",p[i][j]);
}
break;
}
}
if(!ok){
printf("Impossible");
}printf("\n");
}
return 0;
}J
- Meaning of the questions: given n * m matrix, the elements of each column to move forward from the election, each row Select All will get a reward, seeking to get the maximum benefit.
- Idea: enumerate j selected as the shortest column (rewarded), and then enumerate the maximum benefit for each row (selected from the current to the end)
#include<bits/stdc++.h>
#define ll long long
#define FOR(i,l,r) for(int i = l ; i <= r ;++i )
#define inf 1<<30
#define EPS (1e-9)
#define ALL(T) T.begin(),T.end()
#define lson(i) i<<1
#define rson(i) (i<<1|1)
using namespace std;
typedef pair<int,int> pii;
const int maxn = 1e4+10;
ll c[maxn][maxn],d[maxn],mx[maxn][maxn],smx[maxn];
int n,m;
int main(){
int t;
scanf("%d",&t);
FOR(cas,1,t){
scanf("%d%d",&n,&m);
int x;
FOR(i,1,n){
FOR(j,1,m){
scanf("%d",&x);
c[i][j]=c[i][j-1]-x;
}
mx[i][m+1] = -1ll<<60;
for(int j=m;j>=0;--j){
mx[i][j] = max(mx[i][j+1],c[i][j]);
}
}
FOR(j,0,m){
smx[j] = 0;
FOR(i,1,n){
smx[j] += mx[i][j];
}
}
ll ans = 0,D=0,sum;
for(int j=0;j<=m;++j){
if(j) scanf("%d",&x),D+=x;
sum = -1ll<<60;
for(int i=1;i<=n;++i){
// if(c[i][j]-mx[i][j] > sum){
// sum = c[i][j] - mx[i][j];
// }
sum = c[i][j] - mx[i][j] + smx[j];
ans = max(ans,sum+D);
}
// sum+= smx[j];
// for(int i=1;i<=n;++i) sum+= mx[i][j];
// ans = max(ans,sum+D);
}
printf("Case #%d: %lld\n",cas,ans);
}
return 0;
}