https://scut.online/p/337

这个东西是个阶梯状的。那么可以考虑存两棵树，一棵树是阶梯的，另一棵树的平的，随便一减就是需要的阶梯。

优化之后貌似速度比树状数组还惊人。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;inline int read() {int x=0;int f=0;char c;do {c=getchar();if(c=='-')f=1;} while(c<'0'||c>'9');do {x=(x<<3)+(x<<1)+c-'0';c=getchar();} while(c>='0'&&c<='9');return f?-x:x;
}inline void _write(int x) {if(x>9)_write(x/10);putchar(x%10+'0');
}inline void write(int x) {if(x<0) {putchar('-');x=-x;}_write(x);putchar('\n');
}void TestCase(int ti);int main() {
#ifdef Yinkufreopen("Yinku.in","r",stdin);//freopen("Yinku.out","w",stdout);
#endif // Yinkuint T=1;for(int ti=1; ti<=T; ti++)TestCase(ti);
}/*---  ---*/const int MAXM=100000;
int a[MAXM+5];
int st[(MAXM<<2)+5];
int st2[(MAXM<<2)+5];const int mod=1000000007;inline int add(const int &a,const int &b){int res=a+b;return res>=mod?res-mod:res;
}inline int sub(const int &a,const int &b){int res=a-b;return res<0?res+mod:res;
}inline int mul(const int &a,const int &b){ll res=1ll*a*b;return res>=mod?res%mod:res;
}inline void push_up(int o) {st[o]=add(st[o<<1],st[o<<1|1]);st2[o]=add(st2[o<<1],st2[o<<1|1]);
}void build(int o,int l,int r) {if(l==r){st[o]=a[l];st2[o]=mul(l,a[l]);}else {int m=(l+r)>>1;build(o<<1,l,m);build(o<<1|1,m+1,r);push_up(o);}
}void update(int o,int l,int r,int x,int v) {if(l==r) {st[o]=v;st2[o]=mul(x,v);return;} else {int m=(l+r)>>1;if(x<=m)update(o<<1,l,m,x,v);else if(x>=m+1)update(o<<1|1,m+1,r,x,v);push_up(o);}
}int query1(int o,int l,int r,int a,int b) {if(a<=l&&r<=b) {return st[o];} else {int m=(l+r)>>1;int ans=0;if(a<=m)ans=query1(o<<1,l,m,a,b);if(b>=m+1)ans=add(ans,query1(o<<1|1,m+1,r,a,b));return ans;}
}int query2(int o,int l,int r,int a,int b) {if(a<=l&&r<=b) {return st2[o];} else {int m=(l+r)>>1;int ans=0;if(a<=m)ans=query2(o<<1,l,m,a,b);if(b>=m+1)ans=add(ans,query2(o<<1|1,m+1,r,a,b));return ans;}
}inline void TestCase(int ti) {int n,m;while(~scanf("%d",&n)) {for(int i=1; i<=n; i++)scanf("%d",&a[i]);build(1,1,n);scanf("%d",&m);for(int i=1; i<=m; i++) {char a[2];int b,c;scanf("%s%d%d",a,&b,&c);if(a[0]=='Q') {printf("%d\n",sub(query2(1,1,n,b,c),mul(b-1,query1(1,1,n,b,c))));} else {update(1,1,n,b,c);}}}
}

这个树状数组就不太好懂了，不过空间是线段树的1/4，速度是其两倍。
单点改值就把差值update上去就可以了。然后记得把原始值也顺手改了。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;inline int read() {int x=0;int f=0;char c;do {c=getchar();if(c=='-')f=1;} while(c<'0'||c>'9');do {x=(x<<3)+(x<<1)+c-'0';c=getchar();} while(c>='0'&&c<='9');return f?-x:x;
}inline void _write(int x) {if(x>9)_write(x/10);putchar(x%10+'0');
}inline void write(int x) {if(x<0) {putchar('-');x=-x;}_write(x);putchar('\n');
}void TestCase(int ti);int main() {
#ifdef Yinkufreopen("Yinku.in","r",stdin);//freopen("Yinku.out","w",stdout);
#endif // Yinkuint T=1;for(int ti=1; ti<=T; ti++)TestCase(ti);
}/*---  ---*/const int MAXM=100000;
int a[MAXM+5];
int bit[MAXM+5];
int bit2[MAXM+5];int n;const int mod=1000000007;inline int add(const int &a,const int &b) {int res=a+b;return res>=mod?res-mod:res;
}inline int sub(const int &a,const int &b) {int res=a-b;return res<0?res+mod:res;
}inline int mul(const int &a,const int &b) {ll res=1ll*a*b;return res>=mod?res%mod:res;
}inline int sum(int x,int bit[]) {int res=0;while(x) {res=add(res,bit[x]);x-=x&-x;}return res;
}inline void update(int x,int v,int bit[]) {while(x<=n) {bit[x]=add(bit[x],v);x+=x&-x;}
}inline int range_sum(int x,int y,int bit[]) {return sub(sum(y,bit),sum(x-1,bit));
}inline void TestCase(int ti) {int m;while(~scanf("%d",&n)) {memset(bit,0,sizeof(bit));memset(bit2,0,sizeof(bit2));for(int i=1; i<=n; i++) {scanf("%d",&a[i]);update(i,a[i],bit);update(i,mul(i,a[i]),bit2);}scanf("%d",&m);for(int i=1; i<=m; i++) {char s[2];int b,c;scanf("%s%d%d",s,&b,&c);if(s[0]=='Q') {printf("%d\n",sub(range_sum(b,c,bit2),mul(b-1,range_sum(b,c,bit))));} else {int delta=sub(c,a[b]);update(b,delta,bit);update(b,mul(b,delta),bit2);a[b]=c;}}}
}