ARC 075

C
01背包
D
每次攻击，相当于对全体造成B的伤害，选一个怪兽造成A-B的伤害，二分次数x （二分的上界不要定高了不然会爆longlongQWQ

E
让b[i]=a[i]-k,相当于问b有多少个连续子序列和>=0
（1：枚举右端点，维护每个左端点的答案，可以手写splay每次看<=一个数的左端点有多少个 或 离线树状数组
（2：也可以维护b的前缀和转为二维偏序）

code：

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long long#define lowbit(x) x&(-x)using namespace std;const int maxn = 210000;int n,K;int a[maxn];struct node{int t,k;ll x;}s[maxn<<1]; int sn;inline bool cmp(const node x,const node y){return x.x==y.x?x.k>y.k:x.x>y.x;}int tr[maxn];void add(int x){for(;x<=n;x+=lowbit(x))tr[x]++;}int query(int x){int re=0;for(;x;x-=lowbit(x))re+=tr[x];return re;}ll ans=0;int main(){    scanf("%d%d",&n,&K); for(int i=1;i<=n;i++) scanf("%d",&a[i]),a[i]-=K;    ll now=0;    s[++sn]=(node){0,0,0};    for(int i=1;i<=n;i++)    {        now+=a[i];        s[++sn]=(node){i,0,-now};        s[++sn]=(node){i,1,-now+a[i]};    }sort(s+1,s+sn+1,cmp);    for(int i=1;i<=sn;i++)    {        if(s[i].k) add(s[i].t);        else ans+=query(s[i].t);    }    printf("%lld\n",ans);    return 0;}

F
写出rev(N)=x1x2x3..xn,N=xnxn-1xn-2..x1
rev(N)−N=(x1−xn)∗10n−1,(x2−xn−1)∗10n−2......
容易发现N的位数不超过D的2倍，枚举N的位数L后，从低位到高位，2L/2枚举(xi-xj)的差值，对称过去判是否合法，若合法算一下这种差值的方案数
细节有点多

code：

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long longusing namespace std;const int maxn = 22;ll pw[maxn];int D,dn,d[maxn];int n,a[maxn];ll ans;void calc(){    for(int i=1;i*2<=n;i++) a[i]=-a[n-i+1];    ll now=0;    for(int i=1;i<=n;i++) now+=pw[n-i]*a[i];    if(now==(ll)D)    {        //printf("%d ",n); for(int i=1;i<=n;i++) printf("%d ",a[i]); puts("");        ll num=1ll;        if(n&1) num*=10ll;        for(int i=1;i*2<=n;i++)        {            int tmp=abs(a[i]);            if(i==1) tmp++;            num*=(ll)(10-tmp);        }        ans+=num;    }}void dfs(const int now,const int las){    if(!(n&1)&&now<=n/2) { calc();return; }    if((n&1)&&now==(n+1)/2) { a[now]=0;calc();return; }    int tmp=(d[now-(n-dn)]+las)%10;    a[now]=tmp; dfs(now-1,las&(!tmp));    if(tmp!=0) a[now]=-(10-tmp),dfs(now-1,1);}int main(){    pw[0]=1ll;for(int i=1;i<=18;i++) pw[i]=pw[i-1]*10ll;    scanf("%d",&D);    int tmp=D;while(tmp) d[++dn]=tmp%10,tmp/=10;    for(int i=1;i*2<=dn;i++) swap(d[i],d[dn-i+1]);    for(n=dn;n<=2*dn;n++)         dfs(n,0);    printf("%lld\n",ans);    return 0;}