早
T1
T1的话正解我也不知道是什么,但是呢,暴力能拿分,我手里页没有好的std,将就一下吧。
#include <cstdio>
#include <cstring>
char np[2022][13][32];
int ym[2][14]={
{0,31,28,31,30,31,30,31,31,30,31,30,31,31},
{0,31,29,31,30,31,30,31,31,30,31,30,31,31}
};
int isLeap(int yy){
return (yy%400==0)||(yy%4==0&&yy%100!=0)?1:0;
}
void nextday(int &yy,int &mm,int &dd){
if(mm==12&&dd==31){
yy++;mm=1;dd=1;
return;
}
dd++;
if(dd>ym[isLeap(yy)][mm]){
dd=1;
mm++;
}
}
bool hasTheSameDayOfNextMonth(int yy,int mm,int dd){
return dd<=ym[isLeap(yy)][mm+1];
}
bool dfs(const int yy,const int mm,const int dd){
if(np[yy][mm][dd]!=-1)return np[yy][mm][dd];
if(
yy>2012
|| (yy==2012&&mm==12&&dd>22)
)
return np[yy][mm][dd]=true;
if(yy==2012&&mm==12&&dd==22)
return np[yy][mm][dd]=false;
if(hasTheSameDayOfNextMonth(yy,mm,dd)){
if(mm<12){
if(!dfs(yy,mm+1,dd)){
return np[yy][mm][dd]=true;
}
}else{
if(!dfs(yy+1,1,dd)){
return np[yy][mm][dd]=true;
}
}
}
int ty=yy,tm=mm,td=dd;
nextday(ty,tm,td);
if(!dfs(ty,tm,td))
return np[yy][mm][dd]=true;
return np[yy][mm][dd]=false;
}
int main(){
freopen("calendar.in","r",stdin);
freopen("calendar.out","w",stdout);
int yy,mm,dd;
memset(np,-1,sizeof(np));
while(~scanf("%d%d%d",&yy,&mm,&dd)){
printf("%s\n",dfs(yy,mm,dd)?"YES":"NO");
}
fclose(stdin); fclose(stdout);
return 0;
}
T2
T2的话树状数组加暴力枚举,这个页没什么好说的,还是暴力嘛,我还是不会打。
我只有80P
#include<map>
#include<queue>
#include<cmath>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define qread(x) x=read()
#define mes(x,y) memset(x,y,sizeof(x))
#define mpy(x,y) memcpy(x,y,sizeof(x))
#define Maxn 500000
#define INF 2147483647
inline int read(){
char ch=getchar();
int f=1,x=0;
while(!(ch>='0'&&ch<='9')){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}
return x*f;
}
int n,m,x,y,z,sum[Maxn+1];
int main(){
freopen("gcd.in","r",stdin);
freopen("gcd.out","w",stdout);
qread(n);qread(m);
mes(sum,0);z=-INF;
for(int i=1;i<=n;i++){
qread(x);y=sqrt(x);
for(int j=1;j<=y;j++){
if(x%j==0){
sum[j]++;
sum[x/j]++;
}
}
if(x==y*y)sum[y]--;
z=std::max(z,x);
}
for(int i=z;i>=1;i--){
if(sum[i]>=m){
printf("%d\n",i*m);
return 0;
}
}
}
/*
3 1
1 2 3
*/
std
#include <cstdio>
int n, k, x, m, cnt[500010];
int main(){
freopen("gcd.in", "r", stdin); freopen("gcd.out", "w", stdout);
for (scanf("%d%d", &n, &k); n--;) scanf("%d", &x), cnt[x]++, m = m < x ? x : m;
for (int i = m, tmp; i; i--){
tmp = 0;
for (int j = i; j <= m; j += i) tmp += cnt[j];
if (tmp >= k){
printf("%I64d\n", (long long)i * k);
fclose(stdin); fclose(stdout);
return 0;
}
}
}
T3
T3其实还是暴力,我不会,就是把可以到达最前面且可以拉前来的就拉过来,贪心嘛。
#include<map>
#include<queue>
#include<cmath>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define qread(x) x=read()
#define mes(x,y) memset(x,y,sizeof(x))
#define mpy(x,y) memcpy(x,y,sizeof(x))
#define Maxn 100000
#define INF 2147483647
inline int read(){
char ch=getchar();
int f=1,x=0;
while(!(ch>='0'&&ch<='9')){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}
return x*f;
}
int n,m,tmp,len[2];
char s[Maxn+1];
int main(){
freopen("pasuwado.in","r",stdin);
freopen("pasuwado.out","w",stdout);
qread(n);m=0;
scanf("%s",s+1);len[0]=strlen(s+1);
while(n>0&&m<len[0]){
len[1]=std::min(n+m+1,len[0]);
tmp=m+1;
for(int i=m+2;i<=len[1];i++)if(s[tmp]>s[i])tmp=i;
n-=tmp-m-1;
s[0]=s[tmp];
for(int i=tmp;i>=m+2;i--)s[i]=s[i-1];
s[m+1]=s[0];
m++;
}
s[len[0]+1]='\0';printf("%s\n",s+1);
}
/*
3
dcba
*/
std,std只有pas我很无奈啊。
var
pk,tmp,now,tt,i,j,k:longint;
n,m:int64;
ans,s:ansistring;
r:array['a'..'z']of longint;
t:array[1..100000]of longint;
w:array[1..100000,1..2]of longint;
p:array[1..100000]of boolean;
ch:char;
procedure ins(x:longint);
begin
if x>n then exit;
inc(t[x]);
ins(x+x and (-x));
end;
function found(x:longint):longint;
begin
if x=0 then exit(0);
exit(t[x]+found(x-x and (-x)));
end;
procedure add(x:char;y:longint);
begin
inc(tt);
w[tt,1]:=y;
w[tt,2]:=r[x];
r[x]:=tt;
end;
begin
assign(input,'pasuwado.in'); reset(input);
assign(output,'pasuwado.out'); rewrite(output);
readln(m);
readln(s);
n:=length(s);
for i:=n downto 1 do add(s[i],i);
for now:=1 to n do begin
for ch:='a' to 'z' do begin
if r[ch]=0 then continue;
tmp:=w[r[ch],1]-1-found(w[r[ch],1]);
if tmp<=m then begin
m:=m-tmp;
ins(w[r[ch],1]);
ans:=ans+s[w[r[ch],1]];
r[ch]:=w[r[ch],2];
break;
end;
end;
end;
writeln(ans);
close(input); close(output);
end.
查看原文:http://hz2016.cn/blog/?p=121