Codeforces Round #578 (Div. 2)

Codeforces Round #578 (Div. 2)

A

# include <bits/stdc++.h>
using namespace std;

int r[15];
int main()
{
int n;
scanf("%d",&n);
getchar();
char a;
for(int j=0;j<n;j++){
scanf("%c",&a);
if(a=='L'){
for(int i=0;i<10;i++){
if(r[i]==0){
r[i]=1;
break;
}
}
}else if(a=='R'){
for(int i=9;i>=0;i--){
if(r[i]==0){
r[i]=1;
break;
}
}
}else{
r[a-'0']=0;
}
}
for(int i=0;i<10;i++){
printf("%d",r[i]);
}


return 0;
}

B

# include <bits/stdc++.h>
using namespace std;

const int MAXN=110;
int h[MAXN];
int main()
{
int t;
scanf("%d",&t);
while(t--){
int n,m,k;
int flag=1;
scanf("%d %d %d",&n,&m,&k);
scanf("%d",&h[1]);
for(int i=2;i<=n;i++){
scanf("%d",&h[i]);
int low=h[i]-k;
low=max(0,low);
m+=(h[i-1]-low);
if(m<0) flag=0;
}
if(flag) printf("YES\n");
else printf("NO\n");
}

return 0;
}

C

# include <bits/stdc++.h>
using namespace std;

typedef long long LL;
LL gcd(LL a,LL b)
{
return b==0?a:gcd(b,a%b);
}
int main()
{
LL n,m,q;
scanf("%lld %lld %lld",&n,&m,&q);
LL sx,sy,ex,ey;
LL gd=gcd(n,m);
LL nn=n/gd,mm=m/gd;
for(int i=0;i<q;i++){
scanf("%lld %lld %lld %lld",&sx,&sy,&ex,&ey);
if(sx==1){
sy=(sy-1)/nn;
}else{
sy=(sy-1)/mm;
}
if(ex==1){
ey=(ey-1)/nn;
}else{
ey=(ey-1)/mm;
}
if(sy==ey) printf("YES\n");
else printf("NO\n");
}

return 0;
}



posted @ 2022-02-26 22:53  fengzlj  阅读(148)  评论(0)    收藏  举报