A Simple Problem with Integers
Time Limit: 5000MS | Memory Limit: 131072KB | 64bit IO Format: %I64d & %I64u |
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 abc" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q ab" 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
抄的大神的模板,注意懒惰标记的下传
代码:
#include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #include<string> #include<cmath> #include<queue> #include<vector> #include<map> #include<set> #define INF 0x3f3f3f3f #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; const int maxd=100000+5; const int maxn=10000+5; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1 | 1 typedef long long ll; typedef pair<int,int> pii; //--------------------------- int q,n; ll add[maxd<<2],sum[maxd<<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; 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) { add[rt]+=c; sum[rt]+=(ll) c*(r-l+1); return; } pushdown(rt,r-l+1); 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 ret=0; if(L<=m) ret+=query(L,R,lson); if(m<R) ret+=query(L,R,rson); return ret; } int main() { int kase; freopen("1.txt","r",stdin); while(scanf("%d%d",&n,&q)!=EOF) { build(1,n,1); while(q--) { char op[5]; int a,b,c; scanf("%s",op); if(op[0]=='C') { scanf("%d%d%d",&a,&b,&c); update(a,b,c,1,n,1); } else { scanf("%d%d",&a,&b); printf("%lld\n",query(a,b,1,n,1)); } } } return 0; }
时间: 2024-12-25 19:50:34