3212: Pku3468 A Simple Problem with Integers
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 2046 Solved: 892
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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.
水题 区间加法 区间求和
mmp我又忘了update后pushup
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #define ls u<<1 #define rs ls|1 #define ll long long #define N 100050 using namespace std; int a[N],n,m;ll sum[N<<2],lz[N<<2]; void pushup(int u){sum[u]=sum[ls]+sum[rs];} void build(int u,int l,int r){ if(l==r){ sum[u]=a[l]; return; } int mid=l+r>>1; build(ls,l,mid); build(rs,mid+1,r); pushup(u); } void pushdown(int u,int L,int R){ if(!lz[u])return; int mid=L+R>>1; lz[ls]+=lz[u];lz[rs]+=lz[u]; sum[ls]+=lz[u]*(mid-L+1); sum[rs]+=lz[u]*(R-mid); lz[u]=0; } ll query(int u,int L,int R,int l,int r){ if(l<=L&&R<=r)return sum[u]; pushdown(u,L,R); int mid=L+R>>1;ll t=0; if(l<=mid)t+=query(ls,L,mid,l,r); if(r>mid)t+=query(rs,mid+1,R,l,r); return t; } void update(int u,int L,int R,int l,int r,int val){ if(l<=L&&R<=r){ sum[u]+=val*(R-L+1); lz[u]+=val;return; } pushdown(u,L,R);int mid=L+R>>1; if(l<=mid)update(ls,L,mid,l,r,val); if(r>mid)update(rs,mid+1,R,l,r,val); pushup(u); } int main(){ //freopen(".in","r",stdin); //freopen(".out","w",stdout); scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) scanf("%d",&a[i]); build(1,1,n); char s[10];int l,r,v; while(m--){ scanf("%s%d%d",s,&l,&r); if(s[0]=='Q'){ printf("%lld",query(1,1,n,l,r)); if(m)putchar('\n'); } else scanf("%d",&v),update(1,1,n,l,r,v); } return 0; }