8.20 北校 複習

==。

Div2 區間和

線段樹+二分

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 100010;
map<LL,int> mark;
int tot;
struct Tree{
    int l,r;
    int sum;
    int lazy;
}tree[4*maxn];

void build(int node){
    tree[node].sum = 0;
    if(tree[node].l == tree[node].r) return;
    int mid = (tree[node].l + tree[node].r) /2;
    tree[node*2].l = tree[node].l;
    tree[node*2].r = mid;
    tree[node*2+1].l = mid+1;
    tree[node*2+1].r = tree[node].r;
    build(2*node);
    build(2*node+1);
    return;
}

void tree_insert(int node,int pos){
    if(tree[node].l == pos && tree[node].r == pos){
        tree[node].sum++; return;
    }
    int mid=(tree[node].l+tree[node].r)/2;
    if(pos<=mid){
        tree_insert(2*node,pos);
    }
    else
        tree_insert(2*node+1,pos);

    tree[node].sum = tree[node*2].sum + tree[node*2+1].sum;
    return;
}

int query(int node,int s,int e){
    if(tree[node].l>=s&&tree[node].r<=e){
        //printf("s = %d  e = %d ans = %d\n",s,e,tree[node].sum);
        return tree[node].sum;
    }
    int ans = 0;
    int mid = (tree[node].l + tree[node].r)/2;
    if(s<=mid) ans+=query(node*2,s,e);
    if(e>mid) ans+=query(node*2+1,s,e);
    //printf("s = %d  e = %d ans = %d\n",s,e,ans);
    return ans;
}

LL pre[maxn];
LL line[maxn];

int main(){
    //freopen("1.txt","r",stdin);
    int n,L,R;
    while(~scanf("%d%d%d",&n,&L,&R)){
        tot = 0;
        LL ans = 0;

        for(int i=0;i<n;i++){
            int a; scanf("%d",&a);
            pre[i+1] = a + pre[i];
            line[i] = pre[i+1];
            if(pre[i+1]>=L && pre[i+1]<=R) ans++;
            //printf("pre = %d\n", pre[i+1]);
        }
        sort(line,line+n);
        for(int i=0;i<n;i++)
            if(!mark.count(line[i])) mark[line[i]] = ++tot;

        tree[1].l=1, tree[1].r=tot;
        build(1);

        for(int i=1;i<=n;i++){
            LL* n1 = lower_bound(line,line+n,pre[i]-R);
            LL* n2 = upper_bound(line, line+n, pre[i]-L);
            n2--;
            if(n2>=n1){
                int id1 = mark[*n1];
                int id2 = mark[*n2];
                ans+=query(1,id1,id2);
                //printf("    ans = %lld\n",ans);
            }
            tree_insert(1,mark[pre[i]]);
        }

        printf("%lld\n", ans);
    }
    return 0;
}

Div2 更新學號

#include <bits/stdc++.h>
using namespace std;
int mark[9999999];
int id[5010];

int main(){
    int n;
    while(~scanf("%d", &n)){
        for(int i=0; i<n; i++)
            scanf("%d", &id[i]);
        sort(id,id+n);
        memset(mark, 0, sizeof(mark));
        for(int i=0;i<n;i++){
            for(int j=i+1; j<n; j++)
                mark[id[j]-id[i]]=1;
        }
        int ans = n;
        bool flag=false;
        while(!flag){
            flag = true;
            for(int i=ans; i<1e7; i+=ans){
                if(mark[i]){
                    //printf("i = %d\n", i);
                    flag = false;
                    break;
                }
            }
            ans++;
        }
        printf("%d\n", ans-1);
    }
    return 0;
}

碼題

Div1 wmq的A+B Problem

Div1 Triangles

Div1 wmq的征程