You are given a simple task. Given a sequence A[i] with N numbers. You have to perform Q operations on the given sequence.
Here are the operations:
- A v l, add the value v to element with index l.(1<=V<=1000)
- R a l r, replace all the elements of sequence with index i(l<=i<= r) with a(1<=a<=10^6) .
- Q l r, print the number of elements with index i(l<=i<=r) and A[i] is a prime number
Note that no number in sequence ever will exceed 10^7.
Input
The first line is a signer integer T which is the number of test cases.
For each test case, The first line contains two numbers N and Q (1 <= N, Q <= 100000) - the number of elements in sequence and the number of queries.
The second line contains N numbers - the elements of the sequence.
In next Q lines, each line contains an operation to be performed on the sequence.
Output
For each test case and each query,print the answer in one line.
Sample Input
1 5 10 1 2 3 4 5 A 3 1 Q 1 3 R 5 2 4 A 1 1 Q 1 1 Q 1 2 Q 1 4 A 3 5 Q 5 5 Q 1 5
Sample Output
2 1 2 4 0 4
来自 <http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3911>
【题意】：
单点修改，区间修改，查询区间素数个数
【解题思路】：
维护线段树，结点内容为：Val-实时数值(只针对单点) Sum-区间素数个数 Lazy-区间修改后传递给子区间
关键点在于数值和素数标记的同时更新和传递，由于区间更新时没有直接更新到最底层，故单点查询也需要pushdown操作。这里选择将区间更新后的值直接下传，至于素数标记，下传后再进行判断并赋值。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 #define mid(a,b) ((a+b)>>1) 7 #define LL int 8 #define maxn 110000 9 #define IN freopen("in.txt","r",stdin); 10 using namespace std; 11 12 char is_prime[maxn*100]; 13 void sieve() 14 { 15 int m=(int)sqrt((maxn*100)+0.5); 16 fill(is_prime,is_prime+(maxn*100),1); 17 is_prime[0]=is_prime[1]=0; 18 for(int i=2;i<=m;i++) if(is_prime[i]) 19 for(int j=i*i;j<(maxn*100);j+=i) is_prime[j]=0; 20 } 21 22 int n,q; 23 LL num[maxn]; 24 struct Tree 25 { 26 int left,right; 27 LL sum; 28 LL val,lazy; 29 }tree[maxn<<2]; 30 31 32 /*递归建树*/ 33 void build(int i,int left,int right) 34 { 35 tree[i].left=left; 36 tree[i].right=right; 37 tree[i].lazy=0; 38 39 if(left==right){ 40 tree[i].val=num[left]; 41 tree[i].sum=(is_prime[num[left]]? 1:0); 42 return ; 43 } 44 45 int mid=mid(left,right); 46 47 build(i<<1,left,mid); 48 build(i<<1|1,mid+1,right); 49 50 tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum; 51 } 52 53 /*区间修改，标记下传：每当访问到当前结点的子节点时，下传标记*/ 54 void pushdown(int i) 55 { 56 if(tree[i].lazy){ 57 int tmp = (is_prime[tree[i].lazy]? 1:0); 58 tree[i<<1].val=tree[i].lazy; 59 tree[i<<1|1].val=tree[i].lazy; 60 tree[i<<1].lazy=tree[i].lazy; 61 tree[i<<1|1].lazy=tree[i].lazy; 62 tree[i<<1].sum=(tree[i<<1].right-tree[i<<1].left+1)*tmp; 63 tree[i<<1|1].sum=(tree[i<<1|1].right-tree[i<<1|1].left+1)*tmp; 64 tree[i].lazy=0; /*下传后清零*/ 65 } 66 } 67 68 /*区间修改，d为改变量*/ 69 void update(int i,int left,int right,LL d) 70 { 71 if(tree[i].left==left&&tree[i].right==right) 72 { 73 int tmp = (is_prime[d]? 1:0); 74 tree[i].sum=(right-left+1)*tmp; 75 tree[i].val=d; 76 tree[i].lazy=d; 77 return ; 78 } 79 80 pushdown(i); 81 82 int mid=mid(tree[i].left,tree[i].right); 83 84 if(right<=mid) update(i<<1,left,right,d); 85 else if(left>mid) update(i<<1|1,left,right,d); 86 else 87 { 88 update(i<<1,left,mid,d); 89 update(i<<1|1,mid+1,right,d); 90 } 91 92 tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum; 93 } 94 95 /*单点修改，d为改变量*/ 96 void update(int i,int x,LL d) 97 { 98 if(tree[i].left==tree[i].right){ 99 tree[i].val+=d; 100 tree[i].sum=(is_prime[tree[i].val]? 1:0); 101 return; 102 } 103 104 pushdown(i); 105 int mid=mid(tree[i].left,tree[i].right); 106 107 if(x<=mid) update(i<<1,x,d); 108 else update(i<<1|1,x,d); 109 110 tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum; 111 } 112 113 /*区间结果查询*/ 114 LL query(int i,int left,int right) 115 { 116 if(tree[i].left==left&&tree[i].right==right) 117 return tree[i].sum; 118 119 pushdown(i); 120 121 int mid=mid(tree[i].left,tree[i].right); 122 123 if(right<=mid) return query(i<<1,left,right); 124 else if(left>mid) return query(i<<1|1,left,right); 125 else return query(i<<1,left,mid)+query(i<<1|1,mid+1,right); 126 } 127 128 129 int main(int argc, char const *argv[]) 130 { 131 //IN; 132 133 sieve(); 134 int t;scanf("%d",&t); 135 while(t--) 136 { 137 scanf("%d %d",&n,&q); 138 for(int i=1;i<=n;i++) scanf("%d",&num[i]); 139 build(1,1,n); 140 141 while(q--) 142 { 143 char c; 144 while(c=getchar()){ 145 if(c=='A'||c=='R'||c=='Q') break; 146 } 147 if(c=='A'){ 148 int v,l; 149 scanf("%d %d",&v,&l); 150 update(1,l,v); 151 } 152 if(c=='R'){ 153 int a,l,r; 154 scanf("%d %d %d",&a,&l,&r); 155 update(1,l,r,a); 156 } 157 if(c=='Q'){ 158 int l,r; 159 scanf("%d %d",&l,&r); 160 printf("%d ", query(1,l,r)); 161 } 162 } 163 } 164 165 return 0; 166 }