poj 3468 A Simple Problem with Integers
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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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4
Sample Output
455915
线段树的区间更新,用到了延时更新标记,很难调啊
#include <stdio.h>#define N 100001#define ll long long#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1ll add[N<<2];ll sum[N<<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]+=(m-(m>>1))*add[rt]; sum[rt<<1|1]+=(m>>1)*add[rt]; add[rt]=0; }}void Build(int l,int r,int rt){ add[rt]=0; if(l==r) { scanf("%I64d",&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(R>m) 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(R>m) ret+=Query(L,R,rson); return ret;}int main(){ int m,n; scanf("%d%d",&n,&m); Build(1,n,1); while(m--) { char s[5]; int a,b,c; 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;}
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