标签：tween tip divide input following nta dea 操作 lin
InputThere are no more than 10 test cases.
For each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000.
Each the following m lines contains an operation. Operation 1 to 3 is in this format: "1 x y c" or "2 x y c" or "3 x y c". Operation 4 is in this format: "4 x y p". (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3)
The input ends with 0 0.
OutputFor each operation 4, output a single integer in one line representing the result. The answer may be quite large. You just need to calculate the remainder of the answer when divided by 10007.Sample Input
5 5 3 3 5 7 1 2 4 4 4 1 5 2 2 2 5 8 4 3 5 3 0 0
Sample Output
307 7489
题意：就是三种操作，一种加上一个数，一种乘上一个数，一种改为一个数，然后就是求最终一段区间每个数的几次方，相加。
题解：就是有一个基础值，然后就是存储其三次方，二次方，一次方的值，这样便于最后的输出，这个是十分十分复杂的。
如果在赛场上需要保持十分良好的心态才可以AC，三次方，二次方，一次方的更新，不是特别难，但是需要十分细心，以及
良好的耐心。
1 #include<cstdio> 2 #include<algorithm> 3 #include<iostream> 4 #include<cmath> 5 #define lson(x) x*2 6 #define rson(x) x*2+1 7 using namespace std; 8 const int MAXN=200007; 9 const int mod=10007; 10 typedef long long LL; 11 int n,m,x,y,z,num; 12 struct hh{ 13 int l,r,mid,add,mul,zhi[3]; 14 }tree[MAXN<<2]; 15 void push_up(int p) 16 { 17 for (int i=0;i<3;i++) 18 tree[p].zhi[i]=(tree[lson(p)].zhi[i]+tree[rson(p)].zhi[i])%mod; 19 } 20 void finish_work(int p,int mul,int add) 21 { 22 int len=tree[p].r-tree[p].l+1; 23 tree[p].zhi[0]=tree[p].zhi[0]*mul%mod; 24 tree[p].zhi[1]=tree[p].zhi[1]*mul%mod*mul%mod; 25 tree[p].zhi[2]=tree[p].zhi[2]*mul%mod*mul%mod*mul%mod; 26 27 tree[p].mul=tree[p].mul*mul%mod; 28 tree[p].add=(tree[p].add*mul%mod+add)%mod; 29 30 tree[p].zhi[2]=(tree[p].zhi[2]+3*add%mod*add%mod*tree[p].zhi[0]%mod)%mod; 31 tree[p].zhi[2]=(tree[p].zhi[2]+3*add%mod*tree[p].zhi[1]%mod)%mod; 32 tree[p].zhi[2]=(tree[p].zhi[2]+len*add%mod*add%mod*add%mod)%mod; 33 tree[p].zhi[1]=(tree[p].zhi[1]+2*add%mod*tree[p].zhi[0]%mod)%mod; 34 tree[p].zhi[1]=(tree[p].zhi[1]+len*add%mod*add%mod)%mod; 35 tree[p].zhi[0]=(tree[p].zhi[0]+len*add%mod)%mod; 36 } 37 void push_down(int p) 38 { 39 if (tree[p].l==tree[p].r) return; 40 finish_work(lson(p),tree[p].mul,tree[p].add); 41 finish_work(rson(p),tree[p].mul,tree[p].add); 42 tree[p].mul=1,tree[p].add=0; 43 } 44 void build(int l,int r,int p) 45 { 46 tree[p].l=l,tree[p].r=r,tree[p].mid=(l+r)>>1,tree[p].mul=1; 47 tree[p].add=tree[p].zhi[1]=tree[p].zhi[2]=tree[p].zhi[0]=0; 48 if (l==r) return; 49 build(l,tree[p].mid,lson(p)); 50 build(tree[p].mid+1,r,rson(p)); 51 } 52 void change(int l,int r,int p,int mul,int add) 53 { 54 //cout<<l<<" "<<r<<" "<<p<<endl; 55 if (l<=tree[p].l&&r>=tree[p].r) finish_work(p,mul,add); 56 else { 57 push_down(p); 58 if (r<=tree[p].mid) change(l,r,lson(p),mul,add); 59 else if (l>tree[p].mid) change(l,r,rson(p),mul,add); 60 else{ 61 change(l,tree[p].mid,lson(p),mul,add); 62 change(tree[p].mid+1,r,rson(p),mul,add); 63 } 64 push_up(p); 65 } 66 } 67 int query(int l,int r,int p,int num) 68 { 69 //cout<<l<<" "<<r<<" "<<p<<endl; 70 if (l<=tree[p].l&&r>=tree[p].r) return tree[p].zhi[num-1]; 71 push_down(p); 72 if (r<=tree[p].mid) return query(l,r,lson(p),num); 73 else if (l>tree[p].mid) return query(l,r,rson(p),num); 74 else return (query(l,tree[p].mid,lson(p),num)+query(tree[p].mid+1,r,rson(p),num))%mod; 75 } 76 int main() 77 { 78 while (scanf("%d%d",&n,&m)&&(n+m)) 79 { 80 build(1,n,1); 81 for (int i=1;i<=m;i++) 82 { 83 scanf("%d%d%d%d",&num,&x,&y,&z); 84 if (num==1) change(x,y,1,1,z); 85 else if (num==2) change(x,y,1,z,0); 86 else if (num==3) change(x,y,1,0,z); 87 else printf("%d\n",query(x,y,1,z)%mod); 88 } 89 } 90 }
标签：tween tip divide input following nta dea 操作 lin
原文地址：http://www.cnblogs.com/fengzhiyuan/p/7535839.html
