Codeforces Round #552 (Div. 3)
休闲VP发现打的真的太休闲了,完全没有比赛的氛围
A,B
不管
C
暴力跑第一轮和最后一轮
中间的轮数直接做除法
#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<ctime>
#include<cmath>
#define LL long long
#define pii pair<int,int>
#define mk make_pair
#define fi first
#define se second
using namespace std;
inline int read(){
int v = 0,c = 1;char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') c = -1;
ch = getchar();
}
while(isdigit(ch)){
v = v * 10 + ch - 48;
ch = getchar();
}
return v * c;
}
LL aaa,bbb,ccc;
LL ans;
int aa[5],bb[5],cc[5];
inline LL work(LL x,LL a,LL b,LL c){
for(int j = x;j <= 7;++j){
if(j == 1 || j == 4 || j == 7) a--;
if(j == 2 || j == 6) b--;
if(j == 3 || j == 5) c--;
if(a < 0 || b < 0 || c < 0) return j - x;
}
LL week,day;
LL resta = a / 3;
LL restb = b / 2;
LL restc = c / 2;
week = min(resta,min(restb,restc));
LL ans = week * 7 + 7 - x + 1;
a = a - week * 3,b -= week * 2,c -= week * 2;
for(int j = 1;j <= 7;++j){
if(j == 1 || j == 4 || j == 7) a--;
if(j == 2 || j == 6) b--;
if(j == 3 || j == 5) c--;
ans++;
if(a < 0 || b < 0 || c < 0) return ans - 1;
}
for(int j = 1;j <= 7;++j){
if(j == 1 || j == 4 || j == 7) a--;
if(j == 2 || j == 6) b--;
if(j == 3 || j == 5) c--;
ans++;
if(a < 0 || b < 0 || c < 0) return ans - 1;
}
}
int main(){
cin >> aaa >> bbb >> ccc;
aa[1] = 4,aa[2] = 7,aa[3] = 8;
bb[1] = 6,bb[2] = 9;
cc[1] = 5,cc[2] = 10;
for(int i = 1;i <= 7;++i){
ans = max(ans,work(i,aaa,bbb,ccc));
// printf("%d %d
",i,ans);
}
cout << ans;
return 0;
}
D
之前贪心发现贪错了
发现操作分位3种
(x-1),(y - 1),(x-1,y+1)
可以证明当(s[i] = 1)时,使用(x - 1,y+1)一定不会更劣
所以就尽可能的在(s[i] = 1)时使用(x)就好了
#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<ctime>
#include<cmath>
#define LL long long
#define pii pair<int,int>
#define mk make_pair
#define fi first
#define se second
using namespace std;
const int N = 2e5 + 3;
int s[N];
int n,a,b,num;//b：太阳能
inline int read(){
int v = 0,c = 1;char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') c = -1;
ch = getchar();
}
while(isdigit(ch)){
v = v * 10 + ch - 48;
ch = getchar();
}
return v * c;
}
int main(){
n = read(),a = read(),b = read(),num = 0;
int rest = b;
for(int i = 1;i <= n;++i) s[i] = read();
for(int i = 1;i <= n;++i){
if(a == 0 && rest == 0) {printf("%d
",i - 1);return 0;}
if(s[i] == 0 && rest != 0) rest--;
else if(s[i] == 0 && rest == 0) a--;
else if(s[i] == 1 && rest != b){
if(a == 0) rest--;
else {
a--;
if(rest != 0) num++;
rest++;
}
}
else if(s[i] == 1 && rest == b){
rest--;
}
// printf("%d %d
",a,rest);
}
printf("%d
",n);
return 0;
}
/*
19 7 4
0 0 1 1 0 1 0 0 1 0 1 1 1 1 1 1 0 0 1
*/
E
直接用双向链表模拟就好了
时间复杂度可以证明
暴力删除
懒得写链表,就用set维护
#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<ctime>
#include<set>
#include<cmath>
#define LL long long
#define pii pair<int,int>
#define mk make_pair
#define fi first
#define se second
using namespace std;
const int N = 2e5 + 3;
int a[N];
int L[N];
int R[N];
bool book[N];
int n,k;
int ans[N];
set<pii> s;
set<int> ss;
int sta[N];
int tot;
set<pii>::iterator it;
set<int>::iterator iit;
inline int read(){
int v = 0,c = 1;char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') c = -1;
ch = getchar();
}
while(isdigit(ch)){
v = v * 10 + ch - 48;
ch = getchar();
}
return v * c;
}
int main(){
int now = 1;
n = read(),k = read();
for(int i = 1;i <= n;++i) a[i] = read(),s.insert(mk(a[i],i)),ss.insert(i);
while(!s.empty()){
it = s.end();it--;
pii x = *it;
tot = 0;
iit = ss.lower_bound(x.se);
for(int num = 0;num <= k;--iit,++num){
// printf("%d ",*iit);
ans[*iit] = now;
s.erase(mk(a[*iit],*iit));
sta[++tot] = *iit;
if(iit == ss.begin()) break;
}
iit = ss.lower_bound(x.se);
for(int num = 0;num <= k && iit != ss.end();++iit,++num){
// printf("%d ",*iit);
ans[*iit] = now;
s.erase(mk(a[*iit],*iit));
sta[++tot] = *iit;
}
// puts("");
for(int i = 1;i <= tot;++i) ss.erase(sta[i]);
if(now == 1) now = 2;
else now = 1;
}
for(int i = 1;i <= n;++i) printf("%d",ans[i]);
return 0;
}
F
发现(k)这么小,肯定有东西
发现购买的物品一定是前(k)
前(k)小一定是从大往小买
直接排序
所以我们设(f_i)表示前(i)个物品的最小花费
每次枚举用了哪一个优惠
或者不用优惠就好了
#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<ctime>
#include<cmath>
#define LL long long
#define pii pair<int,int>
#define mk make_pair
#define fi first
#define se second
using namespace std;
const int N = 2e5 + 3;
inline int read(){
int v = 0,c = 1;char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') c = -1;
ch = getchar();
}
while(isdigit(ch)){
v = v * 10 + ch - 48;
ch = getchar();
}
return v * c;
}
int n,k,m;
int bar[N];
int f[N];
int a[N];
int sum[N];
int main(){
n = read(),m = read(),k = read();
for(int i = 1;i <= n;++i) a[i] = read();
for(int i = 1;i <= m;++i){
int x = read(),y = read();
bar[x] = max(bar[x],y);
}
sort(a + 1,a + n + 1);
reverse(a + 1,a + k + 1);
for(int i = 1;i <= k;++i) sum[i] = sum[i - 1] + a[i];
memset(f,0x3f,sizeof(f));
f[0] = 0;
// for(int i = 1;i <= k;++i) printf("%d
",a[i]);
for(int i = 1;i <= k;++i){
for(int j = 1;j <= i;++j)
f[i] = min(f[i],f[i - j] + sum[i] - sum[i - j] - (sum[i] - sum[i - bar[j]]));
f[i] = min(f[i],f[i - 1] + a[i]);
}
cout << f[k] << endl;
return 0;
}
G
[lcm(i,j) = frac{i imes j}{gcd(i,j)}
]
我们就枚举(gcd(i,j))
显然(i,j)都要是他的倍数
在枚举的时候顺带更新答案就好了
CF上开(4)s,(10^7)一个(log)还是可以过得
#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<vector>
#include<ctime>
#include<cmath>
#define LL long long
#define pii pair<int,int>
#define mk make_pair
#define fi first
#define se second
using namespace std;
const int N = 1e7 + 3;
const int M = 1e7;
int n;
int a[N];
LL ans = 9999999999999999;
pii id;
int vis[N];
inline int read(){
int v = 0,c = 1;char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') c = -1;
ch = getchar();
}
while(isdigit(ch)){
v = v * 10 + ch - 48;
ch = getchar();
}
return v * c;
}
int main(){
n = read();
for(int i = 1;i <= n;++i){
a[i] = read();
if(vis[a[i]]){
if(a[i] < ans) ans = a[i],id = mk(vis[a[i]],i);
}
vis[a[i]] = i;
}
for(int d = 1;d <= M;++d){
int maxx1 = 0,maxx2 = 0;
for(int x = d;x <= M;x += d){
if(vis[x]){
if(!maxx1) maxx1 = vis[x];
else if(!maxx2) maxx2 = vis[x];
else maxx2 = maxx1,maxx1 = vis[x];
if(maxx1 != 0 && maxx2 != 0)
if(1ll * a[maxx1] * a[maxx2] / d < ans) ans = 1ll * a[maxx1] * a[maxx2] / d,id = mk(maxx1,maxx2);
// printf("%d %d %d %lld
",d,maxx1,maxx2,ans);
}
}
}
if(id.fi > id.se) swap(id.fi,id.se);
printf("%d %d
",id.fi,id.se);
return 0;
}