CodeForces121E 线段树上线段果

http://codeforces.com/problemset/problem/121/E

题意： Petya 喜欢幸运数，幸运数只包含 4 和 7 这两个数字。例如 47，744，4 都是幸运数字，但 5,16,467 不是。

Petya 有一个 N 个数的数组，他想给这个数组执行 M 个操作，可以分为两种操作：

1. add l r d 把第 l 到 第 r 个数都加上 d；
2. count l r 统计第 l 到第 r 个数有多少个幸运数字。

喜闻乐见的数据结构题。

更加喜闻乐见的是这题能用树状数组直接暴力跑过去。对每一个区间修改进行n次单点修改的傻狗操作竟然也可以AC这题

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Scl(x) scanf("%lld",&x);
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
const double eps = 1e-;
const int maxn = 1e5 + ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
int N,M,tmp,K;
const int sp[] = {,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,};
bool check[];
int a[maxn];
int tree[maxn];
void add(int t,int x){
    for(;t <= N;t += t & -t){
        tree[t] += x;
    }
}
int getsum(int t){
    int s = ;
    for(;t;t -= t & -t){
        s += tree[t];
    }
    return s;
}
int main()
{
    For(i,,) check[sp[i]] = ;
    scanf("%d%d",&N,&M);
    For(i,,N){
        int x; Sca(x); a[i] = x;
        if(check[x]) add(i,);
    }
    For(i,,M){
        int l,r;
        char op[];
        scanf("%s%d%d",op,&l,&r);
        if(op[] == 'a'){
            int d; Sca(d);
            For(j,l,r){
                if(check[a[j]]) add(j,-);
                a[j] += d;
                if(check[a[j]]) add(j,);
            }
        }else{
            Pri(getsum(r) - getsum(l - ));
        }
    }
    #ifdef VSCode
    system("pause");
    #endif
    return ;
}

当然正解肯定不是这样的

困难之处在于维护区间内幸运数字的个数。事实上确实不简单，思路和hdu多校里面的一道线段树相似。

线段树维护三个值，区间内所有数到下一个幸运数字差值的最小值，最小值的个数，lazy标记

每一次update之后要重新遍历一次线段树去寻找有没有最小值小于0的数，然后将这些减少到0的数一个个修改即可。

看起来很麻烦的操作事实上时间复杂度是很科学的正解。

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Scl(x) scanf("%lld",&x);
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
inline LL read(){LL now=;register char c=getchar();for(;!isdigit(c);c=getchar());
for(;isdigit(c);now=now*+c-'',c=getchar());return now;}
inline void out(int x){if(x > ) out(x / ); putchar(x %  + '');}
inline void out(LL x){if(x > ) out(x / ); putchar(x %  + '');}
typedef vector<int> VI;
const double eps = 1e-;
const int maxn = 1e5 + ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
int N,M,tmp,K;
const int sp[] = {,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,};
struct Tree{
    int l,r;
    int x,y,lazy;
}tree[maxn * ];
int c[maxn];
int a[maxn];
void Pushup(int t){
    if(tree[t << ].x == tree[t <<  | ].x){
        tree[t].x = tree[t << ].x;
        tree[t].y = tree[t << ].y + tree[t <<  | ].y;
    }else if(tree[t << ].x < tree[t <<  | ].x){
        tree[t].x = tree[t << ].x;tree[t].y = tree[t << ].y;
    }else{
        tree[t].x = tree[t <<  | ].x;tree[t].y = tree[t <<  | ].y;
    }
}
void Build(int t,int l,int r){
    tree[t].l = l; tree[t].r = r;
    tree[t].lazy = ;
    if(l == r){
        tree[t].x = c[a[l]];
        tree[t].y = ;
        return;
    }
    int m = (l + r) >> ;
    Build(t << ,l,m); Build(t <<  | ,m + ,r);
    Pushup(t);
}
void Pushdown(int t){
    if(tree[t].lazy){
        tree[t << ].lazy += tree[t].lazy;
        tree[t <<  | ].lazy += tree[t].lazy;
        tree[t << ].x += tree[t].lazy;
        tree[t <<  | ].x += tree[t].lazy;
        tree[t].lazy = ;
    }
}
void update(int t,int l,int r,int val){
    if(l <= tree[t].l && tree[t].r <= r){
        tree[t].x += val;
        tree[t].lazy += val;
        return;
    }
    Pushdown(t);
    int m = (tree[t].l + tree[t].r) >> ;
    if(r <= m) update(t << ,l,r,val);
    else if(l > m) update(t <<  | ,l,r,val);
    else{
        update(t << ,l,m,val); update(t <<  | ,m + ,r,val);
    }
    Pushup(t);
}

void rebuild(int t){
    if(tree[t].x >= ) return;
    if(tree[t].l == tree[t].r){
        a[tree[t].l] -= tree[t].lazy;
        tree[t].x = c[a[tree[t].l]];
        tree[t].lazy = ;
        return ;
    }
    Pushdown(t);
    rebuild(t << ); rebuild(t <<  | );
    Pushup(t);
}
int query(int t,int l,int r){
    if(tree[t].x) return ;
    if(l <= tree[t].l && tree[t].r <= r){
        return tree[t].y;
    }
    Pushdown(t);
    int m = (tree[t].l + tree[t].r) >> ;
    if(r <= m) return query(t << ,l,r);
    if(l > m) return query(t <<  | ,l,r);
    else return query(t << ,l,m) + query(t <<  | ,m + ,r);
}
int main()
{
    int cnt = ;
    For(i,,1e4){
        if(sp[cnt] < i) cnt++;
        c[i] = sp[cnt] - i;
    }
    N = read(); M = read();
    For(i,,N) a[i] = read();
    Build(,,N);
    int l,r;
    char op[];
    For(i,,M){
        scanf("%s",op);
        l = read(); r = read();
        if(op[] == 'a'){
            int d; d = read();
            update(,l,r,-d);
            rebuild();
        }else{
            out(query(,l,r));
            puts("");
        }
    }
    #ifdef VSCode
    system("pause");
    #endif
    return ;
}