AtCoder Beginner Contest 174 总结
翘了实时比赛，今天用virtual补一下......发现题变水了......怕是错失了一次上分机会。
A,B,C略。D题每次交换尽可能远的一对"W......R"。E题二分答案，注意下边界初始化不能直接设成0，否则二分判断时除数太小精度爆炸。F虽然n还不小，但还是可以用莫队水过去（理论上
大于3e9，2000ms有点危险），正解应该是主席树或者离线+树状数组。主席树空间开大点没毛病......
A
#include
#include
#include
#include
using namespace std;
typedef long long ll;
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
int main() {
int x=read();
puts(x<30?"No":"Yes");
return 0;
} B
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
int n;
int ans;
double x,y,d;
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
int main() {
n=read();
scanf("%lf",&d);
d*=d;
while (n--) {
scanf("%lf%lf",&x,&y);
ans+=(x*x+y*y<=d);
}
cout< C
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int N=1e6+4;
int m,vis[N],x;
int tim;
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
int main() {
m=read();
while (1) {
++tim;
x=(x*10+7)%m;
if (!x) {
printf("%d\n",tim);
break;
}
if (vis[x]) {
puts("-1");
break;
}
vis[x]=true;
}
return 0;
} D
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int N=2e5+4;
int n;
char s[N];
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
int ans;
int main() {
n=read();
scanf("%s",s+1);
int l=1,r=n;
while (l1) --r;
if (l E
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int N=2e5+4;
int n;
ll m;
double a[N];
double l=1e9,r=0;
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
template inline void smax(T &x,T y) {
x=x inline void smin(T &x,T y) {
x=x>y?y:x;
}
inline bool ok(double x) {
ll cnt=0;
for (register int i=1;i<=n;++i) {
cnt+=(ll)ceil(a[i]/x)-1;
if (cnt>m) return false;
}
return true;
}
int main() {
n=read(),m=read();
for (register int i=1;i<=n;++i) {
scanf("%lf",&a[i]);
smax(r,a[i]);
smin(l,a[i]/(m+1));
}
while (r-l>1e-2) {
double mid=(l+r)/2;
if (ok(mid)) r=mid;
else l=mid;
}
cout<<(ll)ceil(l)< F
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int N=5e5+4;
int n,m;
int root[N],sum[N*50],lc[N*50],rc[N*50];
int tim;
int nxt[N];
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
inline void pushup(int rt) {
sum[rt]=sum[lc[rt]]+sum[rc[rt]];
}
inline void build(int &rt,int l,int r) {
rt=++tim;
if (l==r) return ;
int mid=l+r>>1;
build(lc[rt],l,mid);
build(rc[rt],mid+1,r);
}
inline void modify(int &rt,int pre,int l,int r,int pos,int val) {
rt=++tim;
lc[rt]=lc[pre];
rc[rt]=rc[pre];
sum[rt]=sum[pre]+val;
if (l==r) return ;
int mid=l+r>>1;
if (pos<=mid) modify(lc[rt],lc[pre],l,mid,pos,val);
else modify(rc[rt],rc[pre],mid+1,r,pos,val);
pushup(rt);
}
inline int query(int rt,int l,int r,int L) {
if (l==r) return sum[rt];
int mid=l+r>>1;
if (mid>=L) return sum[rc[rt]]+query(lc[rt],l,mid,L);
else return query(rc[rt],mid+1,r,L);
}
int main() {
// freopen("f3.txt","r",stdin);
n=read(),m=read();
build(root[0],1,n);
for (register int i=1;i<=n;++i) {
int x=read();
if (!nxt[x]) {
modify(root[i],root[i-1],1,n,i,1);
} else {
int temp;
modify(temp,root[i-1],1,n,nxt[x],-1);
modify(root[i],temp,1,n,i,1);
}
nxt[x]=i;
}
while (m--) {
int x=read(),y=read();
printf("%d\n",query(root[y],1,n,x));
}
return 0;
}