https://ac.nowcoder.com/acm/contest/3003/J
题解：
1 #include<bits/stdc++.h> 2 typedef long long ll; 3 using namespace std; 4 const ll mod = 1e9+7; 5 const int maxn = 2e5+5; 6 struct segT{ 7 ll l,r; 8 ll dat; 9 }t1[maxn*4],t2[maxn*4]; //两棵线段树 10 ll k[maxn],b[maxn]; 11 ll ans; 12 void build1(ll p,ll l,ll r){ 13 t1[p].l = l,t1[p].r = r; 14 if(l == r) { t1[p].dat = k[l]%mod;return;} 15 ll mid = (l+r)/2; 16 build1(p*2,l,mid); 17 build1(p*2+1,mid+1,r); 18 t1[p].dat = ( t1[p*2].dat * t1[p*2+1].dat) %mod ; 19 } 20 void build2(ll p,ll l,ll r){ 21 t2[p].l = l,t2[p].r = r; 22 if(l == r) { t2[p].dat = b[l]%mod;return;} 23 ll mid = (l+r)/2; 24 build2(p*2,l,mid); 25 build2(p*2+1,mid+1,r); 26 t2[p].dat =( (t1[p*2+1].dat * t2[p*2].dat)%mod+ t2[p*2+1].dat )%mod ; 27 } 28 void upd1(ll p,ll L,ll R,ll v){ 29 if(t1[p].l == L &&t1[p].r == R ) {t1[p].dat = v;return;} 30 int mid = (t1[p].l + t1[p].r )/2; 31 if (L<=mid) upd1(p*2,L,R,v); 32 else upd1(p*2+1,L,R,v); 33 t1[p].dat = ( t1[p*2].dat * t1[p*2+1].dat )%mod; 34 } 35 void upd2(ll p,ll L,ll R,ll v){ 36 if(t2[p].l == L&&t2[p].r == R ) {t2[p].dat = v;return;} 37 int mid = (t2[p].l + t2[p].r )/2; 38 if(L<=mid) upd2(p*2,L,R,v); 39 else upd2(p*2+1,L,R,v); 40 t2[p].dat =( (t1[p*2+1].dat * t2[p*2].dat)%mod+ t2[p*2+1].dat )%mod ; 41 } 42 43 void query(ll p,ll l,ll r){ 44 if(l<=t2[p].l && r>=t2[p].r ) { 45 ans = (ans*t1[p].dat + t2[p].dat)%mod; //前面的区间*后面区间的t1[p].dat + 后面区间的t2[p].dat 46 return ; 47 } 48 int mid = (t2[p].l + t2[p].r )/2; 49 if(l<=mid) query(p*2,l,r); 50 if(r>mid) query(p*2+1,l,r); 51 } 52 int main() 53 { 54 int n,m; 55 scanf("%d%d",&n,&m); 56 for(int i = 1;i<=n;i++) { 57 scanf("%lld",&k[i]); 58 } 59 for(int i = 1;i<=n;i++){ 60 scanf("%lld",&b[i]); 61 } 62 build1(1,1,n); 63 build2(1,1,n); 64 while(m--){ 65 int f; 66 scanf("%d",&f); 67 if(f == 1) { 68 ll pos,tk,tb; 69 scanf("%lld%lld%lld",&pos,&tk,&tb); 70 upd1(1,pos,pos,tk); 71 upd2(1,pos,pos,tb); 72 // printf("de %d %d " ,t1[1].dat,t2[1].dat); 73 } 74 if(f == 2){ 75 int l,r; 76 scanf("%lld%lld",&l,&r); 77 ans = 1; 78 query(1,l,r); 79 printf("%lld ",ans%mod); 80 } 81 } 82 return 0; 83 }