标签：acm算法 cstring printf namespace lines
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 70482 | Accepted: 21735 | |
| Case Time Limit: 2000MS | ||
Description
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.
Input
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.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
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 4
Sample Output
4 55 9 15
Hint
The sums may exceed the range of 32-bit integers.
线段树区间求和，刚刚理解，，还是理解的不够透彻。。以后有空补上。。
#include <cstdio> #include <algorithm> using namespace std; #define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 #define LL __int64 const int maxn = 111111; LL add[maxn<<2]; LL sum[maxn<<2]; void pushup(int rt) { sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } void pushdown(int rt,int m) { if(add[rt]) { add[rt<<1]+=add[rt]; add[rt<<1|1]+=add[rt]; sum[rt<<1]+=add[rt]*(m-(m>>1)); sum[rt<<1|1]+=add[rt]*(m>>1); add[rt]=0; } } void build(int l,int r,int rt) { add[rt] = 0; //清空操作 if (l == r) { scanf("%lld",&sum[rt]); return ; } int m = (l + r) >> 1; build(lson); build(rson); pushup(rt); } void update(int L,int R,int c,int l,int r,int rt) { if (L <= l && r <= R) //查询 { add[rt] += c; //储存更新操作，占时不对子节点进行操作 sum[rt] += (LL)c * (r - l + 1); return ; } pushdown(rt , r - l + 1); //对之前延时更新进行对应处理 int m = (l + r) >> 1; if (L <= m) update(L , R , c , lson); if (m < R) update(L , R , c , rson); pushup(rt); } LL query(int L,int R,int l,int r,int rt) { if(L<=l &&r<=R) { return sum[rt]; } pushdown(rt , r - l + 1); int m=(l+r)>>1; LL ans=0; if(L<=m) ans+=query(L,R,lson); if(m<R) ans+=query(L,R,rson); return ans; } int main() { int n,q; int a,b,c; while(scanf("%d%d",&n,&q)!=EOF) { build(1,n,1); char s[2]; while(q--) { scanf("%s",s); if(s[0]=='Q') { scanf("%d%d",&a,&b); printf("%I64d\n",query(a,b,1,n,1)); } else { scanf("%d%d%d",&a,&b,&c); update(a,b,c,1,n,1); } } } return 0; }
PKU 3468 A Simple Problem with Integers
标签：acm算法 cstring printf namespace lines
原文地址：http://blog.csdn.net/sky_miange/article/details/45048869
