Time Limit:5000MS Memory Limit:131072KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 3468
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 }
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 }