维护一次幂二次幂三次幂的三个区间和 再维护乘法加法赋值的三个lazy标记
几个幂次和之间是有公式的 如图
而lazy标记的使用也需要技巧
首先赋值的优先级最高 只要赋值操作 其他两种操作直接舍弃 而加法和乘法虽然优先级一样 但是乘法操作不止影响乘法lazy 也影响加法lazy 具体见代码
#include <bits/stdc++.h> using namespace std; #define ll long long #define M 10007 struct node { int l; int r; ll val1; ll val2; ll val3; ll laz1; ll laz2; ll laz3; }; node tree[400010]; int n; void changeI(int cur,ll val) { ll len; len=tree[cur].r-tree[cur].l+1; tree[cur].val3=(tree[cur].val3+3*val*tree[cur].val2+3*val*val*tree[cur].val1+val*val*val*len)%M; tree[cur].val2=(tree[cur].val2+2*val*tree[cur].val1+val*val*len)%M; tree[cur].val1=(tree[cur].val1+val*len)%M; return; } void changeII(int cur,ll val) { tree[cur].val1=(tree[cur].val1*val)%M; tree[cur].val2=(tree[cur].val2*val*val)%M; tree[cur].val3=(tree[cur].val3*val*val*val)%M; return; } void changeIII(int cur,ll val) { ll len; len=tree[cur].r-tree[cur].l+1; tree[cur].val1=(val*len)%M; tree[cur].val2=(((val*val)%M)*len)%M; tree[cur].val3=(((val*val)%M)*((val*len)%M))%M; return; } void pushup(int cur) { tree[cur].val1=(tree[2*cur].val1+tree[2*cur+1].val1)%M; tree[cur].val2=(tree[2*cur].val2+tree[2*cur+1].val2)%M; tree[cur].val3=(tree[2*cur].val3+tree[2*cur+1].val3)%M; return; } void pushdown(int cur) { if(tree[cur].laz3!=-1) { changeIII(2*cur,tree[cur].laz3); tree[2*cur].laz3=tree[cur].laz3; changeIII(2*cur+1,tree[cur].laz3); tree[2*cur+1].laz3=tree[cur].laz3; tree[2*cur].laz1=0,tree[2*cur+1].laz1=0; tree[2*cur].laz2=1,tree[2*cur+1].laz2=1; tree[cur].laz3=-1; } if(tree[cur].laz1!=0||tree[cur].laz2!=1) { //tree[2*cur].laz1=(tree[2*cur].laz1*tree[cur].laz2+tree[cur].laz1)%M; changeII(2*cur,tree[cur].laz2); changeI(2*cur,tree[cur].laz1); tree[2*cur].laz2=(tree[2*cur].laz2*tree[cur].laz2)%M; tree[2*cur].laz1=(tree[2*cur].laz1*tree[cur].laz2+tree[cur].laz1)%M; changeII(2*cur+1,tree[cur].laz2); changeI(2*cur+1,tree[cur].laz1); tree[2*cur+1].laz2=(tree[2*cur+1].laz2*tree[cur].laz2)%M; tree[2*cur+1].laz1=(tree[2*cur+1].laz1*tree[cur].laz2+tree[cur].laz1)%M; tree[cur].laz1=0; tree[cur].laz2=1; } return; } void build(int l,int r,int cur) { int m; tree[cur].l=l; tree[cur].r=r; tree[cur].val1=0; tree[cur].val2=0; tree[cur].val3=0; tree[cur].laz1=0; tree[cur].laz2=1; tree[cur].laz3=-1; if(l==r) return; m=(l+r)/2; build(l,m,2*cur); build(m+1,r,2*cur+1); return; } void update(int op,int pl,int pr,ll val,int cur) { if(pl<=tree[cur].l&&tree[cur].r<=pr) { if(op==1) { changeI(cur,val); tree[cur].laz1=(tree[cur].laz1+val)%M; } else if(op==2) { changeII(cur,val); tree[cur].laz2=(tree[cur].laz2*val)%M; tree[cur].laz1=(tree[cur].laz1*val)%M; } else { changeIII(cur,val); tree[cur].laz1=0,tree[cur].laz2=1,tree[cur].laz3=val; } return; } pushdown(cur); if(pl<=tree[2*cur].r) update(op,pl,pr,val,2*cur); if(pr>=tree[2*cur+1].l) update(op,pl,pr,val,2*cur+1); pushup(cur); return; } ll query(int tp,int pl,int pr,int cur) { ll res; if(pl<=tree[cur].l&&tree[cur].r<=pr) { if(tp==1) return tree[cur].val1; else if(tp==2) return tree[cur].val2; else return tree[cur].val3; } pushdown(cur); res=0; if(pl<=tree[2*cur].r) res=(res+query(tp,pl,pr,2*cur))%M; if(pr>=tree[2*cur+1].l) res=(res+query(tp,pl,pr,2*cur+1))%M; return res; } int main() { ll val; int q,op,l,r; while(1) { scanf("%d%d",&n,&q); if(n==0&&q==0) break; build(1,n,1); while(q--) { scanf("%d%d%d%lld",&op,&l,&r,&val); if(op!=4) update(op,l,r,val,1); else printf("%lld\n",query(val,l,r,1)%M); } } return 0; }