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：
add l r d把第 l 到 第 r 个数都加上 d；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 ", x) #define Prl(x) printf("%lld ",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-9; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,tmp,K; const int sp[30] = {4,7,44,47,74,77,444,447,474,477,744,747,774,777,4444,4447,4474,4477,4744,4747,4774,4777,7444,7447,7474,7477,7744,7747,7774,7777,}; bool check[10010]; 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 = 0; for(;t;t -= t & -t){ s += tree[t]; } return s; } int main() { For(i,0,29) check[sp[i]] = 1; scanf("%d%d",&N,&M); For(i,1,N){ int x; Sca(x); a[i] = x; if(check[x]) add(i,1); } For(i,1,M){ int l,r; char op[20]; scanf("%s%d%d",op,&l,&r); if(op[0] == 'a'){ int d; Sca(d); For(j,l,r){ if(check[a[j]]) add(j,-1); a[j] += d; if(check[a[j]]) add(j,1); } }else{ Pri(getsum(r) - getsum(l - 1)); } } #ifdef VSCode system("pause"); #endif return 0; }
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 ", x) #define Prl(x) printf("%lld ",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=0;register char c=getchar();for(;!isdigit(c);c=getchar()); for(;isdigit(c);now=now*10+c-'0',c=getchar());return now;} inline void out(int x){if(x > 9) out(x / 10); putchar(x % 10 + '0');} inline void out(LL x){if(x > 9) out(x / 10); putchar(x % 10 + '0');} typedef vector<int> VI; const double eps = 1e-9; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,tmp,K; const int sp[31] = {4,7,44,47,74,77,444,447,474,477,744,747,774,777,4444,4447,4474,4477,4744,4747,4774,4777,7444,7447,7474,7477,7744,7747,7774,7777,44444}; struct Tree{ int l,r; int x,y,lazy; }tree[maxn * 4]; int c[maxn]; int a[maxn]; void Pushup(int t){ if(tree[t << 1].x == tree[t << 1 | 1].x){ tree[t].x = tree[t << 1].x; tree[t].y = tree[t << 1].y + tree[t << 1 | 1].y; }else if(tree[t << 1].x < tree[t << 1 | 1].x){ tree[t].x = tree[t << 1].x;tree[t].y = tree[t << 1].y; }else{ tree[t].x = tree[t << 1 | 1].x;tree[t].y = tree[t << 1 | 1].y; } } void Build(int t,int l,int r){ tree[t].l = l; tree[t].r = r; tree[t].lazy = 0; if(l == r){ tree[t].x = c[a[l]]; tree[t].y = 1; return; } int m = (l + r) >> 1; Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r); Pushup(t); } void Pushdown(int t){ if(tree[t].lazy){ tree[t << 1].lazy += tree[t].lazy; tree[t << 1 | 1].lazy += tree[t].lazy; tree[t << 1].x += tree[t].lazy; tree[t << 1 | 1].x += tree[t].lazy; tree[t].lazy = 0; } } 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) >> 1; if(r <= m) update(t << 1,l,r,val); else if(l > m) update(t << 1 | 1,l,r,val); else{ update(t << 1,l,m,val); update(t << 1 | 1,m + 1,r,val); } Pushup(t); } void rebuild(int t){ if(tree[t].x >= 0) 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 = 0; return ; } Pushdown(t); rebuild(t << 1); rebuild(t << 1 | 1); Pushup(t); } int query(int t,int l,int r){ if(tree[t].x) return 0; if(l <= tree[t].l && tree[t].r <= r){ return tree[t].y; } Pushdown(t); int m = (tree[t].l + tree[t].r) >> 1; if(r <= m) return query(t << 1,l,r); if(l > m) return query(t << 1 | 1,l,r); else return query(t << 1,l,m) + query(t << 1 | 1,m + 1,r); } int main() { int cnt = 0; For(i,1,1e4){ if(sp[cnt] < i) cnt++; c[i] = sp[cnt] - i; } N = read(); M = read(); For(i,1,N) a[i] = read(); Build(1,1,N); int l,r; char op[20]; For(i,1,M){ scanf("%s",op); l = read(); r = read(); if(op[0] == 'a'){ int d; d = read(); update(1,l,r,-d); rebuild(1); }else{ out(query(1,l,r)); puts(""); } } #ifdef VSCode system("pause"); #endif return 0; }