标签：i++ lin hang bit pad read theme its type
Ryuji is not a good student, and he doesn‘t want to study. But there are n books he should learn, each book has its knowledge a[i]a[i].
Unfortunately, the longer he learns, the fewer he gets.
That means, if he reads books from ll to rr, he will get a[l] * L + a[l+1] *(L-1) + …… + a[r-1] *2 + a[r]（L is the length of [ l， r ] that equals to r?l+1).
Now Ryuji has qq questions, you should answer him:
1. If the question type is 1, you should answer how much knowledge he will get after he reads books [ l, r ].
2. If the question type is 2, Ryuji will change the ith book‘s knowledge to a new value.
First line contains two integers nn and qq (n, q≤100000).
The next line contains n integers represent a[i](a[i]≤1e9) .
Then in next qq line each line contains three integers a, b, c, if a = 1， it means question type is 1, and b, ccrepresents [ l, r ]. if a =2 , it means question type is 2 , and bb, cc means Ryuji changes the bth book‘ knowledge to cc
For each question, output one line with one integer represent the answer.
5 3 1 2 3 4 5 1 1 3 2 5 0 1 4 5
10 8
1 #define ull unsigned long long 2 #define ll long long 3 #define N 100009 4 #define lowbit(x) x&(-x) 5 ull c1[N],c2[N]; 6 int n,q; 7 void update1(int x,ull num) 8 { 9 while(x<=n) 10 { 11 c1[x]+=num; 12 x+=lowbit(x); 13 } 14 } 15 ull getsum1(int x) 16 { 17 ull sum=0; 18 while(x>0) 19 { 20 sum+=c1[x]; 21 x-=lowbit(x); 22 } 23 return sum; 24 } 25 void update2(int x,ull num) 26 { 27 while(x<=n) 28 { 29 c2[x]+=num; 30 x+=lowbit(x); 31 } 32 } 33 ull getsum2(int x) 34 { 35 ull sum=0; 36 while(x>0) 37 { 38 sum+=c2[x]; 39 x-=lowbit(x); 40 } 41 return sum; 42 } 43 int main() 44 { 45 scanf("%d%d",&n,&q); 46 ull x; 47 for(int i=1;i<=n;i++) 48 { 49 scanf("%lld",&x); 50 update1(i,x); 51 update2(i,1ull*(n-i+1)*x); 52 } 53 int op,l,r; 54 while(q--) 55 { 56 scanf("%d%d%d",&op,&l,&r); 57 if(op==1) 58 { 59 ull ans1=1ull*(n-r)*(getsum1(r)-getsum1(l-1)); 60 ull ans2=getsum2(r)-getsum2(l-1); 61 printf("%lld\n",ans2-ans1); 62 } 63 else{ 64 ull temp=getsum1(l)-getsum1(l-1); 65 update1(l,1ull*(r-temp) ); 66 update2(l,1ull*(r-temp)*(n-l+1)) ; 67 } 68 } 69 return 0; 70 } 71 72 73
ACM-ICPC 2018 徐州赛区网络预赛 H. Ryuji doesn't want to study
标签：i++ lin hang bit pad read theme its type
原文地址：https://www.cnblogs.com/tingtin/p/9615071.html
