hdu多校 2

。。。 后面四个小时都在挂机很难受。

1010

裸的逆序对

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=;while(b){if(b&)ans=ans*a%mod;a=a*a%mod,b>>=;}return ans;}

using namespace std;

const double eps=1e-;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=+,maxn=+,inf=0x3f3f3f3f;

struct BIT{
    ll sum[N];
    void add(int i,int v)
    {
        for(;i<N;i+=i&(-i))sum[i]+=v;
    }
    ll query(int i)
    {
        ll ans=;
        for(;i;i-=i&(-i))ans+=sum[i];
        return ans;
    }
}b;
int Hash[N],a[N];
int main()
{
    int n,x,y;
    while(~scanf("%d%d%d",&n,&x,&y))
    {
        memset(b.sum,,sizeof b.sum);
        int cnt=;
        for(int i=;i<=n;i++)
        {
            scanf("%d",&a[i]);
            Hash[cnt++]=a[i];
        }
        sort(Hash,Hash+cnt);
        cnt=unique(Hash,Hash+cnt)-Hash;
        ll ans=;
        for(int i=;i<=n;i++)
        {
            a[i]=lower_bound(Hash,Hash+cnt,a[i])-Hash+;
//            printf("%d ",a[i]);
            b.add(a[i],);
            ans+=b.query(cnt+)-b.query(a[i]);
        }
//        puts("");
        printf("%lld\n",ans*min(x,y));
    }
    return ;
}
/********************

********************/

1004

因为1的存在所以先手必胜

1007

因为序列b为 1, 2, 3, 4, 5, 6 .... n， 所以所有点的更新次数不超过 m/1 + m/2 + m/3 .... + m/n = mlogm，

所以就是暴力扣出要更新的点更新就好啦。 天天写搓lazy，反思一下。。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 1e5 + ;
const int M = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +;

int n, q, b[N], mx[N << ], lazy[N << ], stk[N], tot;
char s[N];

struct BIT {
    int a[N];
    void init() {
        memset(a, , sizeof(a));
    }
    void modify(int x, int v) {
        for(int i = x; i <= n; i += i & -i)
            a[i] += v;
    }
    int query(int x) {
        int ans = ;
        for(int i = x; i; i -= i & -i)
            ans += a[i];
        return ans;
    }
} bit;

void build(int l, int r, int rt) {
    lazy[rt] = ;
    if(l == r) {
        mx[rt] = -b[l];
        return;
    }

    int mid = l + r >> ;
    build(l, mid, rt << );
    build(mid + , r, rt <<  | );
    mx[rt] = max(mx[rt << ], mx[rt <<  | ]);
}

void pushdown(int rt) {
    if(lazy[rt] != ) {
        lazy[rt << ] += lazy[rt];
        lazy[rt <<  | ]  += lazy[rt];
        mx[rt << ] += lazy[rt];
        mx[rt <<  | ] += lazy[rt];
        lazy[rt] = ;
    }
}

void update(int L, int R, int v, int l, int r, int rt) {
    if(l >= L && r <= R) {
        lazy[rt] += v;
        mx[rt] += v;
        return;
    }

    int mid = l + r >> ;

    pushdown(rt);

    if(L <= mid) update(L, R, v, l, mid, rt << );
    if(R > mid) update(L, R, v, mid + , r, rt <<  | );

    mx[rt] = max(mx[rt << ], mx[rt <<  | ]);
}

void solve(int l, int r, int rt) {
    if(mx[rt] < ) return;
    if(l == r) {
        stk[tot++] = l;
        return;
    }

    int mid = l + r >> ;
    pushdown(rt);

    solve(l, mid, rt << );
    solve(mid + , r, rt <<  | );

}

int main() {

    while(scanf("%d%d", &n, &q) != EOF) {
        bit.init();
        for(int i = ; i <= n; i++) {
            scanf("%d", &b[i]);
        }
        build(, n, );

        while(q--) {
            int l, r;
            scanf("%s%d%d", s, &l, &r);
            if(s[] == 'a') {
                update(l, r, , , n, );
                tot = ;
                solve(, n, );

                for(int i = ; i < tot; i++) {
                    update(stk[i], stk[i], -b[stk[i]], , n, );
                    bit.modify(stk[i], );
                }

            } else {
                printf("%d\n", bit.query(r) - bit.query(l - ));
            }
        }
    }
    return ;
}

/*
*/

补题-----------------------------------------------------------------

1005

比赛的时候一直构造不出来。。 很神奇的构造方法。。。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N =  + ;
const int M = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +;

int a[N][N];

int main() {

    int p = ;
    for(int i = ; i < p; i++) {
        for(int j = ; j < p; j++) {
            for(int k = ; k < p; k++) {
                a[i * p + j][k * p + (j * k + i) % p] = ;
            }
        }
    }

    puts("");
    for(int i = ; i < ; i++) {
        for(int j = ; j < ; j++) {
            printf("%d", a[i][j]);
        }
        puts("");
    }
    return ;
}

/*
*/

1003

后面一直在写这个题， 构造虚拟点，去掉多余的边，方法都想出来了，可能是对欧拉回路不够了解，一直没a，后来队友补了。