BZOJ3212 A Simple Problem with Integers

3212: Pku3468 A Simple Problem with Integers

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 1277 Solved: 559
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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.

一开始没看懂是和可能超过32位。。long long+树状数组水过，不知为何BZOJ还有这种水题。。
附上本蒟蒻的代码：

#include<cstdio>using namespace std;int n,m,i,x,y,z,k;long long a[100001],p;char c[2];int lowbit(int x){    return x&(-x);}void add(int loc,int value){    int j;    for (j=loc;j<=n;j+=lowbit(j))      a[j]+=value;}long long query(int loc){    int j;    long long ans=0;    for (j=loc;j>=1;j-=lowbit(j))      ans+=a[j];    return ans;}int main(){    scanf("%d %d",&n,&m);    for (i=1;i<=n;i++)      {        scanf("%lld",&p);        add(i,p);      }    for (i=1;i<=m;i++)      {        scanf("%s",&c);        if (c[0]=='Q')          {            scanf("%d %d",&x,&y);            printf("%lld\n",query(y)-query(x-1));          }        if (c[0]=='C')          {            scanf("%d %d %d",&x,&y,&z);            for (k=x;k<=y;k++)              add(k,z);          }      }    return 0;}