NOIP模拟10.21(洛谷10月R2)

洛谷10月月赛R2
A.浮游大陆的68号岛(模拟+前缀和)AC。注意开ll，分左右情况讨论即可。
B.Chtholly Nota Seniorious 完全不会。水了20分。
B的题解看这里:传送门
C.Nephren Ruq Insania 瞎搞了60分。欧拉定理我哪会啊。。

A

#include <bits/stdc++.h>using namespace std;#define N 200010#define inf 0x3f3f3f3f#define ll long long#define mod 19260817inline int read(){    int x=0,f=1;char ch=getchar();    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,d[N],w[N];ll sw[N],s[N];int main(){//  freopen("sample5.in","r",stdin);//  freopen("a.out","w",stdout);    n=read();m=read();    for(int i=2;i<=n;++i) d[i]=read(),d[i]=(d[i]+d[i-1])%mod;    for(int i=1;i<=n;++i) w[i]=read(),sw[i]=(sw[i-1]+w[i])%mod;    for(int i=1;i<=n;++i) s[i]=(ll)w[i]*d[i]%mod,s[i]=(s[i]+s[i-1])%mod;    while(m--){        int x=read(),l=read(),r=read();ll res=0;        if(x<=l) res=s[r]-s[l-1]-(sw[r]-sw[l-1])*d[x];        else if(x>=r) res=(sw[r]-sw[l-1])*d[x]-(s[r]-s[l-1]);        else res=s[r]-s[x]-(sw[r]-sw[x])*d[x]-(s[x-1]-s[l-1]-(sw[x-1]-sw[l-1])*d[x]);        if(res<0) res=res+mod*10000000000LL;        printf("%lld\n",res%mod);    }return 0;}

B

#include <bits/stdc++.h>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define N 2010inline int read(){    int x=0,f=1;char ch=getchar();    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,a[4][N][N],tmp[N][N],mx=0,mn=inf,now=0;void rotate(int x,int y){    for(int i=1;i<=n;++i)        for(int j=1;j<=m;++j)            a[y][j][n-i+1]=a[x][i][j];}bool check(int x){    int last=m;    for(int i=1;i<=n;++i){        for(int j=1;j<=last;++j)            if(a[now][i][j]<mx-x){last=j-1;break;}        for(int j=last+1;j<=m;++j)            if(a[now][i][j]>mn+x) return 0;    }return 1;}bool jud(int x){    if(check(x)) return 1;now++;now&=3;swap(n,m);    if(check(x)) return 1;now++;now&=3;swap(n,m);    if(check(x)) return 1;now++;now&=3;swap(n,m);    if(check(x)) return 1;return 0;}int main(){//  freopen("sample3.in","r",stdin);    n=read();m=read();    for(int i=1;i<=n;++i)        for(int j=1;j<=m;++j) a[now][i][j]=read(),mx=max(mx,a[now][i][j]),mn=min(mn,a[now][i][j]);    for(int i=1;i<=3;++i){        rotate(now,now+1);now++;swap(n,m);    }now=0;swap(n,m);    int l=0,r=mx-mn;    while(l<=r){        int mid=l+r>>1;        if(jud(mid)) r=mid-1;else l=mid+1;    }printf("%d\n",r+1);    return 0;}

C

瞎搞60分版。

#include <bits/stdc++.h>using namespace std;#define N 100010#define inf 0x3f3f3f3f#define ll long longinline int read(){    int x=0,f=1;char ch=getchar();    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,mod;ll a[N];struct que{    int op,l,r,x;}q[N];struct node{    ll x;}tree[N<<2];void build(int p,int l,int r){    if(l==r){tree[p].x=a[l];return;}    int mid=l+r>>1;    build(p<<1,l,mid);build(p<<1|1,mid+1,r);}void add(int p,int l,int r,int x,int y,int val){    if(x<=l&&r<=y){tree[p].x+=val;return;}    int mid=l+r>>1;    if(x<=mid) add(p<<1,l,mid,x,y,val);    if(y>mid) add(p<<1|1,mid+1,r,x,y,val);}ll ask(int p,int l,int r,int x){    if(l==r) return tree[p].x;    int mid=l+r>>1;    if(x<=mid) return ask(p<<1,l,mid,x)+tree[p].x;    else return ask(p<<1|1,mid+1,r,x)+tree[p].x;}inline ll ksm(ll base,ll k){    ll res=1;    for(;k;k>>=1,base=base*base%mod)        if(k&1) res=res*base%mod;return res;}void solve0(){    while(m--){        int op=read(),l=read(),r=read(),x=read();        if(op==1) for(int i=l;i<=r;++i) a[i]+=x;        if(op==2){            int res=a[r]%x;mod=x;            for(int i=r-1;i>=l;--i) res=ksm(a[i]%x,res);            printf("%d\n",res);        }    }}void solve1(){    for(int i=1;i<=m;++i){        if(q[i].op==1) add(1,1,n,q[i].l,q[i].r,q[i].x);        if(q[i].op==2){            mod=q[i].x;            if(q[i].l==q[i].r) printf("%lld\n",ask(1,1,n,q[i].l)%mod);            else printf("%lld\n",ksm(ask(1,1,n,q[i].l)%mod,ask(1,1,n,q[i].r)));        }    }}void solve2(){    for(int i=1;i<=m;++i){        if(q[i].op==1) add(1,1,n,q[i].l,q[i].r,q[i].x);        if(q[i].op==2){            if(ask(1,1,n,q[i].l)%2) puts("1");            else puts("0");        }    }}int main(){    freopen("sample3.in","r",stdin);    freopen("a.out","w",stdout);    n=read();m=read();for(int i=1;i<=n;++i) a[i]=read();    if(n==5){solve0();return 0;}bool flag1=1,flag2=1;    for(int i=1;i<=m;++i){        q[i].op=read(),q[i].l=read(),q[i].r=read(),q[i].x=read();        if(q[i].op==2&&q[i].r-q[i].l+1>2) flag1=0;        if(q[i].op==2&&q[i].x!=2) flag2=0;    }build(1,1,n);    if(flag1){solve1();return 0;}    if(flag2){solve2();return 0;}    return 0;}