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
思路:坑爹的64位整数啊,终于AC了。
#include <stdio.h> #include <string.h> __int64 sum[300000],add[300000],N,Q,ans; void build(__int64 left,__int64 right,__int64 index) { if(left==right) { scanf("%I64d",&sum[index]); add[index]=sum[index]; } else { __int64 mid=(left+right)>>1; build(left,mid,index<<1); build(mid+1,right,index<<1|1); sum[index]=sum[index<<1]+sum[index<<1|1]; } } void updateroot(__int64 left,__int64 right,__int64 index) { sum[index]=0; if(left<right) sum[index]=sum[index<<1]+sum[index<<1|1]; sum[index]+=add[index]*(right-left+1); } void update(__int64 L,__int64 R,__int64 left,__int64 right,__int64 index,__int64 v) { if(L<=left && right<=R) add[index]+=v; else { __int64 mid=(left+right)>>1; if(L<=mid) update(L,R,left,mid,index<<1,v); if(R>mid) update(L,R,mid+1,right,index<<1|1,v); } updateroot(left,right,index); } void query(__int64 L,__int64 R,__int64 left,__int64 right,__int64 index,__int64 attach) { if(L<=left && right<=R) ans+=sum[index]+attach*(right-left+1); else { __int64 mid=(left+right)>>1; if(L<=mid) query(L,R,left,mid,index<<1,attach+add[index]); if(R>mid) query(L,R,mid+1,right,index<<1|1,attach+add[index]); } } int main() { __int64 i,a,b,c; char s[100]; while(~scanf("%I64d%I64d",&N,&Q)) { memset(sum,0,sizeof(sum)); memset(add,0,sizeof(add)); build(1,N,1); gets(s); for(i=0;i<Q;i++) { gets(s); if(s[0]==‘Q‘) { sscanf(s,"%*s%I64d%I64d",&a,&b); ans=0; query(a,b,1,N,1,0); printf("%I64d\n",ans); } else { sscanf(s,"%*s%I64d%I64d%I64d",&a,&b,&c); update(a,b,1,N,1,c); } } } }
poj2486 Apple Tree(树形DP,有反回的情况,求最大)........很典型,布布扣,bubuko.com