A Simple Problem with Integers 线段树 成段更新

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

Hint

The sums may exceed the range of 32-bit integers.
  1 #include <math.h>
  2 #include <stdio.h>
  3 #include <string.h>
  4 #include <algorithm>
  5 using namespace std;
  6 
  7 int a[100005],sum[450005],lazy[450005],ql,qr,c;
  8 
  9 void down(int o,int l,int r)
 10 {
 11     if(lazy[o]!=0)
 12     {
 13         int mid=(l+r)/2;
 14         lazy[o*2]=lazy[o*2]+lazy[o];
 15         lazy[o*2+1]=lazy[o*2+1]+lazy[o];
 16         sum[o*2]=sum[o*2]+lazy[o]*(mid-l+1);
 17         sum[o*2+1]=sum[o*2+1]+lazy[o]*(r-mid);
 18         lazy[o]=0;
 19     }
 20 }
 21 
 22 void up(int o)
 23 {
 24     sum[o]=sum[o*2]+sum[o*2+1];
 25 }
 26 
 27 void build(int o,int l,int r)
 28 {
 29     if(l==r)
 30     {
 31         sum[o]=a[l];
 32         return;
 33     }
 34     int mid=(l+r)/2;
 35     build(o*2,l,mid);
 36     build(o*2+1,mid+1,r);
 37     up(o);
 38 }
 39 
 40 void update(int o,int l,int r)
 41 {
 42     if(ql<=l && qr>=r)
 43     {
 44         lazy[o]=lazy[o]+c;
 45         sum[o]=sum[o]+c*(r-l+1);
 46         return;
 47     }
 48     down(o,l,r);
 49     int mid=(l+r)/2;
 50     if(ql<=mid)
 51         update(o*2,l,mid);
 52     if(qr>mid)
 53         update(o*2+1,mid+1,r);
 54     up(o);
 55 }
 56 
 57 int query(int o,int l,int r)
 58 {
 59     if(ql<=l && qr>=r)
 60     {
 61         return sum[o];
 62     }
 63     down(o,l,r);
 64     int mid=(l+r)/2,ans=0;
 65     if(ql<=mid)
 66         ans=ans+query(o*2,l,mid);
 67     if(qr>mid)
 68         ans=ans+query(o*2+1,mid+1,r);
 69     up(o);
 70     return ans;
 71 }
 72 
 73 int main()
 74 {
 75     int n,q,i;
 76     char sre[10];
 77     while(scanf("%d %d",&n,&q)!=EOF)
 78     {
 79         memset(sum,0,sizeof(sum));
 80         memset(lazy,0,sizeof(lazy));
 81         for(i=1;i<=n;i++)
 82             scanf("%d",&a[i]);
 83 
 84         build(1,1,n);
 85         for(i=1;i<=q;i++)
 86         {
 87             scanf("%s %d %d",sre,&ql,&qr);
 88             if(sre[0]=='C')
 89             {
 90                 scanf("%d",&c);
 91                 update(1,1,n);
 92             }
 93             else
 94             {
 95                 printf("%d\n",query(1,1,n));
 96             }
 97         }
 98     }
 99     return 0;
100 }
View Code

 64位

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <algorithm>
  4 using namespace std;
  5 
  6 long long a[100005],sum[450005],lazy[450005],c;
  7 int ql,qr;
  8 
  9 void down(int o,int l,int r)
 10 {
 11     if(lazy[o]!=0)
 12     {
 13         int mid=(l+r)/2;
 14         lazy[o*2]=lazy[o*2]+lazy[o];
 15         lazy[o*2+1]=lazy[o*2+1]+lazy[o];
 16         sum[o*2]=sum[o*2]+lazy[o]*(mid-l+1);
 17         sum[o*2+1]=sum[o*2+1]+lazy[o]*(r-mid);
 18         lazy[o]=0;
 19     }
 20 }
 21 
 22 void up(int o)
 23 {
 24     sum[o]=sum[o*2]+sum[o*2+1];
 25 }
 26 
 27 void build(int o,int l,int r)
 28 {
 29     if(l==r)
 30     {
 31         sum[o]=a[l];
 32         return;
 33     }
 34     int mid=(l+r)/2;
 35     build(o*2,l,mid);
 36     build(o*2+1,mid+1,r);
 37     up(o);
 38 }
 39 
 40 void update(int o,int l,int r)
 41 {
 42     if(ql<=l && qr>=r)
 43     {
 44         lazy[o]=lazy[o]+c;
 45         sum[o]=sum[o]+c*(r-l+1);
 46         return;
 47     }
 48     down(o,l,r);
 49     int mid=(l+r)/2;
 50     if(ql<=mid)
 51         update(o*2,l,mid);
 52     if(qr>mid)
 53         update(o*2+1,mid+1,r);
 54     up(o);
 55 }
 56 
 57 long long query(int o,int l,int r)
 58 {
 59     if(ql<=l && qr>=r)
 60     {
 61         return sum[o];
 62     }
 63     down(o,l,r);
 64     int mid=(l+r)/2;
 65     long long ans=0;
 66     if(ql<=mid)
 67         ans=ans+query(o*2,l,mid);
 68     if(qr>mid)
 69         ans=ans+query(o*2+1,mid+1,r);
 70     up(o);
 71     return ans;
 72 }
 73 
 74 int main()
 75 {
 76     int n,q,i;
 77     char sre[10];
 78     while(scanf("%d %d",&n,&q)!=EOF)
 79     {
 80         memset(sum,0,sizeof(sum));
 81         memset(lazy,0,sizeof(lazy));
 82         for(i=1;i<=n;i++)
 83             scanf("%lld",&a[i]);
 84         build(1,1,n);
 85         for(i=1;i<=q;i++)
 86         {
 87             scanf("%s %d %d",sre,&ql,&qr);
 88             if(sre[0]=='C')
 89             {
 90                 scanf("%lld",&c);
 91                 update(1,1,n);
 92             }
 93             else
 94             {
 95                 printf("%lld\n",query(1,1,n));
 96             }
 97         }
 98     }
 99     return 0;
100 }
View Code

 

posted @ 2015-08-19 21:07  cyd2014  阅读(163)  评论(0)  编辑  收藏  举报