平时十五测

题解：

第一题：

或者ＤＰ，先按ｘ排序，ｄｐ［ｉ］表示选择ｉ作为结尾的最大团size;

发现i向前连边的条件是xi - wi >= xj + wj; (这个式子我推出来，但并没有深入探究）

简单探索一番可以发现如果i可以和前面的j连边，j可以和前面p连边，则i就可以和p连边；

dp[i] = max(dp[j]) + 1, 这个可以用树状数组维护；

#include<bits/stdc++.h>
using namespace std;

const int M = 200005;
int c[M<<1], dis[M<<1], q;
struct node{int u, v, x;}p[M];
bool cmp(node A, node B){
    return A.x < B.x;
}
void add(int x, int u){
    for(; x <= q; x += x&(-x)) c[x] = max(c[x], u);
}
int query(int x){
    int ret = 0;
    for(; x; x -= x&(-x)) ret = max(ret, c[x]);
    return ret;
}
int read(){
    int x = 0; int f = 1; char c = getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c<='9'&&c>='0'){x=x*10+c-'0';c=getchar();}
    return x*=f;
}

int main(){
    freopen("clique.in","r",stdin);
    freopen("clique.out","w",stdout);
    int n = read(), tot = 0;
    for(int i = 1; i <= n; i++){
        int x = read(), w = read();
        p[i].x = x;
        p[i].u = x - w;
        p[i].v = x + w; 
        dis[++tot] = p[i].u, dis[++tot] = p[i].v;
    }
    sort(dis + 1, dis + 1 + tot);
    q = unique(dis + 1, dis + 1 + tot) - dis - 1;
    for(int i = 1; i <= n; i++){
        p[i].u = lower_bound(dis + 1, dis + 1 + q, p[i].u) - dis;
        p[i].v = lower_bound(dis + 1, dis + 1 + q, p[i].v) - dis;
    }
    sort(p + 1, p + 1 + n, cmp);
    add(p[1].v, 1);
    int ans = 1;
    for(int i = 2; i <= n; i++){
        int t = query(p[i].u);
        add(p[i].v, t+1);
        ans = max(ans, t+1);    
    }
    printf("%d
", ans);
    
}

第二题：正确解法基于一个重要的结论，a % b <= a/2，这就意味着，任何一个数loga[i]次取模以内就能变为0。同时还注意到，a % b = a (a < b)。这两点便可以极大的降低时间复杂度。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 1e5+5;
int a[M], n, m;
#define Ls nd->ls, lf, mid
#define Rs nd->rs, mid+1, rg
struct Node{
    int v; ll sum;
    Node *ls, *rs;
    void up(){
        v = max(ls->v, rs->v);
        sum = ls->sum + rs->sum;
    }
    
}pool[M << 2], *root, *tail = pool; 
Node *build(int lf = 1, int rg = n){
    Node *nd = ++tail;
    if(lf == rg)nd->v = nd->sum = a[lf];
    else {
        int mid = (lf + rg) >> 1;
        nd->ls = build(lf, mid);
        nd->rs = build(mid+1, rg);
        nd->up();
    }
    return nd;
}
void add(int pos, int x, Node *nd = root, int lf = 1, int rg = n){
    if(lf == rg){
        nd->v = nd->sum = x;
    }
    else {
        int mid = (lf + rg) >> 1;
        if(pos <= mid)add(pos, x, Ls);
        else add(pos, x, Rs);
        nd->up();
    }
}

ll query(int L, int R, Node *nd = root, int lf = 1, int rg = n){
    if(L <= lf && rg <= R)return nd->sum;
    else {
        int mid = (lf + rg) >> 1;
        ll ret = 0;
        if(L <= mid)ret += query(L, R, Ls);
        if(R > mid) ret += query(L, R, Rs);
        return ret;
    }
    
}
void modify(int L, int R, ll x, Node *nd = root, int lf = 1, int rg = n){
    if(nd->v < x) return ;
    if(lf == rg)nd->v %= x, nd->sum%= x;
    else {
        int mid = (lf + rg) >> 1;
        int ret = 0;
        if(L <= mid)modify(L, R, x, Ls);
        if(R > mid) modify(L, R, x, Rs);
        nd->up();
    }
}
int read(){
    int x = 0; int f = 1; char c = getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c<='9'&&c>='0'){x=x*10+c-'0';c=getchar();}
    return x*=f;
}

int main(){
    freopen("mod.in","r",stdin);
    freopen("mod.out","w",stdout);
    n = read(), m = read();
    for(int i = 1; i <= n; i++) a[i] = read();
    root = build();
    int opt, l, r, x;
    while(m--){
        opt = read();
        if(opt == 1){
            l = read(), r = read();
            printf("%lld
", query(l, r));
        }
        else if(opt == 2){
            l = read(), r = read(), x = read();
            modify(l, r, x);
        }
        else {
            l = read(), x = read();
            add(l, x);
        }
    }
    
    
}

第三题：令S(i)={i+1,i+2,…,2i}，f(i,k)表示S(i)中二进制下恰好有k个1的数的个数（i>0，k>=0）。

发现f(i, k) <= f(i+1, k), 这个满足二分性质， 就二分i, f可以用数位DP解决，但是是Log^2的，我们可以直接用组合数解决；

没有顶上界， 有C(位数，要的1)， 代替数位DP的过程，顶上界只有一次，是logN的；

#include<bits/stdc++.h>
using namespace std;
#define ull unsigned long long
const ull INF = 2e18;
ull C[67][67];
int digit[67];
ull dfs(int dep, int res, int sum){
    int now = sum - res;
    if(now > dep)return 0;
    if(!now)return 1;
    if(dep == 1) return digit[dep] == sum - res;
    if(!digit[dep])return dfs(dep-1, res, sum);
    ull tmp = C[dep-1][now];
    if(tmp == INF) return INF + 1;
    ull nw = dfs(dep-1, res+1, sum);
    return min(INF + 1, tmp + nw);
}


ull check(ull xz, ull k){
    ull up = xz<<1;
    int cnt = 0;
    while(xz){
        digit[++cnt] = xz&1;
        xz >>= 1;
    }
    ull ans1 = dfs(cnt, 0, k);
    cnt = 0;
    while(up){
        digit[++cnt] = up&1;
        up >>= 1;
    }
    ull ans2 = dfs(cnt, 0, k);
    return ans2 - ans1;
}


int main(){
    freopen("number.in","r",stdin);
    freopen("number.out","w",stdout);
    for(int i = 0; i <= 65; i++)
        for(int j = 0; j <= i; j++)
            if(j == 0 || i == j) C[i][j] = 1;
            else C[i][j] = min(C[i-1][j-1] + C[i-1][j], INF);
    int T;
    scanf("%d", &T);
    ull m, k;
    while(T--){
        scanf("%llu%llu", &m, &k);
        if(m == 1 && k == 1){puts("1 -1");continue;}
        ull lf = 1, rg = 1e18, p1 = -1, p2 = -1;
        while(lf <= rg){
            ull mid = (lf + rg) / 2;
            ull now = check(mid, k);
            if(now == m){
                p1 = mid;
                rg = mid - 1;
            }
            else if(now > m)rg = mid - 1;
            else lf = mid + 1;
        }
        
        lf = 1, rg = 2e18;
        while(lf <= rg){
            ull mid = (lf + rg) / 2;
            ull now = check(mid, k);
            if(now == m){
                p2 = mid;
                lf = mid + 1;
            }
            else if(now > m)rg = mid - 1;
            else lf = mid + 1;
        }
        
        printf("%llu %llu
", p1, p2 - p1 + 1);
    }
}