kuangbin专题七：C题 A Simple Problem with Integers
Posted 0一叶0知秋0
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了kuangbin专题七：C题 A Simple Problem with Integers相关的知识，希望对你有一定的参考价值。
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
You need to answer all Q commands in order. One answer in a line.
10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4Sample Output
4 55 9 15Hint
The sums may exceed the range of 32-bit integers.
题意：区间累加 ， 区间查询 求和
思路：线段树直接写就好 ， 模板题目 （注意 做好 延迟更新）
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> using namespace std ; #define maxn 110000 #define LL long long LL N , Q ; char ch ; LL a , b , c ; LL num[maxn] ; struct node { LL l , r , add , sum ; }; node tree[maxn*4] ; void build(LL root , LL l , LL r ){ tree[root].l = l ; tree[root].r = r ; tree[root].add = 0 ; if( l == r ){ tree[root].sum = num[l] ; return; } int mid = (l+r)/2 ; build(root*2 , l , mid ) ; build(root*2+1 , mid+1 , r ) ; tree[root].sum = tree[root*2].sum + tree[root*2+1].sum ; } void update(LL root , LL s , LL e , LL value){ if(e < tree[root].l || tree[root].r < s){ return; } if(s<= tree[root].l && tree[root].r <= e){ tree[root].sum += (tree[root].r - tree[root].l+1) * value ; tree[root].add += value ; return; } // 延迟更新 LL add = tree[root].add ; if(add){ tree[root*2].add += add ; tree[root*2].sum += (tree[root*2].r-tree[root*2].l+1) * add ; tree[root*2+1].add += add ; tree[root*2+1].sum += (tree[root*2+1].r-tree[root*2+1].l+1) * add ; tree[root].add = 0 ; } int mid = (tree[root].l + tree[root].r)/2 ; if(e <= mid){ update(root*2 , s , e , value ) ; } else if(mid<s){ update(root*2+1 , s , e , value ) ; } else if( s <= mid&&mid < e ) { update(root*2 , s , mid , value ) ; update(root*2+1 , mid+1 , e , value ) ; } tree[root].sum = tree[root*2].sum + tree[root*2+1].sum ; } LL query(LL root , LL s , LL e){ if(e < tree[root].l || tree[root].r < s){ return 0 ; } if(s<=tree[root].l && tree[root].r <=e){ return tree[root].sum ; return 0 ; } // 延迟更新 LL add = tree[root].add ; if(add){ tree[root*2].add += add ; tree[root*2].sum += (tree[root*2].r-tree[root*2].l+1) * add ; tree[root*2+1].add += add ; tree[root*2+1].sum += (tree[root*2+1].r-tree[root*2+1].l+1) * add ; tree[root].add = 0 ; } LL mid = (tree[root].l+ tree[root].r)/2 ; if(e<=mid){ return query(root*2 , s , e ) ; } else if(mid<s){ return query(root*2+1 , s , e ) ; } else if(s<=mid && mid <e){ return query(root*2 , s, mid ) + query(root*2+1 , mid+1 , e ) ; } } int main(){ while(~scanf("%lld %lld" , &N , &Q)){ for(int i=1 ; i<=N ; i++){ scanf("%lld" , &num[i]) ; } build(1 , 1 , N ) ; for(int i=1 ; i<=Q ; i++){ scanf(" %c" , &ch) ; if( ch == ‘Q‘){ scanf("%lld %lld" , &a , &b) ; printf("%lld\n" , query(1 , a , b )) ; } else if(ch == ‘C‘){ scanf("%lld %lld %lld" , &a , &b , &c) ; update(1 , a , b , c ) ; } } } return 0 ; }
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> using namespace std ; #define maxn 110000 #define LL long long LL N , Q ; char ch ; LL a , b , c ; LL num[maxn] ; LL result ; struct node { LL l , r , add , sum ; }; node tree[maxn*4] ; void build(LL root , LL l , LL r ){ tree[root].l = l ; tree[root].r = r ; tree[root].add = 0 ; if( l == r ){ tree[root].sum = num[l] ; return; } int mid = (l+r)/2 ; build(root*2 , l , mid ) ; build(root*2+1 , mid+1 , r ) ; tree[root].sum = tree[root*2].sum + tree[root*2+1].sum ; } void update(LL root , LL s , LL e , LL value){ if(e < tree[root].l || tree[root].r < s){ return; } if(s<= tree[root].l && tree[root].r <= e){ tree[root].sum += (tree[root].r - tree[root].l+1) * value ; tree[root].add += value ; return; } LL add = tree[root].add ; if(add){ tree[root*2].add += add ; tree[root*2].sum += (tree[root*2].r-tree[root*2].l+1) * add ; tree[root*2+1].add += add ; tree[root*2+1].sum += (tree[root*2+1].r-tree[root*2+1].l+1) * add ; tree[root].add = 0 ; } update(root*2 , s , e , value ) ; update(root*2+1 , s , e , value ) ; tree[root].sum = tree[root*2].sum + tree[root*2+1].sum ; } void query(LL root , LL s , LL e){ if(e < tree[root].l || tree[root].r < s){ return; } if(s<=tree[root].l && tree[root].r <=e){ result += tree[root].sum ; return; } LL add = tree[root].add ; if(add){ tree[root*2].add += add ; tree[root*2].sum += (tree[root*2].r-tree[root*2].l+1) * add ; tree[root*2+1].add += add ; tree[root*2+1].sum += (tree[root*2+1].r-tree[root*2+1].l+1) * add ; tree[root].add = 0 ; } query(root*2 , s , e ) ; query(root*2+1 , s , e ) ; } int main(){ while(~scanf("%lld %lld" , &N , &Q)){ for(int i=1 ; i<=N ; i++){ scanf("%lld" , &num[i]) ; } build(1 , 1 , N ) ; for(int i=1 ; i<=Q ; i++){ scanf(" %c" , &ch) ; if( ch == ‘Q‘){ result = 0 ; scanf("%lld %lld" , &a , &b) ; query(1 , a, b ) ; printf("%lld\n" , result ) ; } else if(ch == ‘C‘){ scanf("%lld %lld %lld" , &a , &b , &c) ; update(1 , a , b , c ) ; } } } return 0 ; }
以上是关于kuangbin专题七：C题 A Simple Problem with Integers的主要内容，如果未能解决你的问题，请参考以下文章
kuangbin专题七 HDU1698 Just a Hook （set线段树）