A Simple Problem with Integers
Time Limit: 5000MSMemory Limit: 131072K
Total Submissions: 77486Accepted: 23862
Case Time Limit: 2000MS
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
Hint
The sums may exceed the range of 32-bit integers.
Source
, Yang Yi
题目链接：?id=3468
题目大意：C情况时a-b区间内每个数字添加c，，Q情况时查询区间内值的和。
解题思路：模板线段树成段更新问题。
代码如下：
#include <cstdio>#include <cstring>#define inf 1e12#define ll long long const ll maxn=1000010;ll n,m,a,b,c;ll sum[maxn],nd[maxn];char s[2];void build(ll l,ll r,ll root){ll mid=(l+r)/2;if(l==r){scanf("%lld",&sum[root]);return;}build(l,mid,root*2);build(mid+1,r,root*2+1);sum[root]=sum[root*2]+sum[root*2+1];}void pushdown(ll l,ll root){if(nd[root]!=0){nd[root*2]+=nd[root];nd[root*2+1]+=nd[root];sum[root*2]+=((l-l/2)*nd[root]);sum[root*2+1]+=((l/2)*nd[root]);nd[root]=0;}}ll query(ll l,ll r,ll root){if(l>=a&&r<=b)return sum[root];pushdown(r-l+1,root);ll mid=(l+r)/2,ans=0;if(a<=mid)ans+=query(l,mid,root*2);if(b>mid)ans+=query(mid+1,r,root*2+1);return ans;}void update(ll l,ll r,ll root){if(l>b||r<a)return;ll len=r-l+1;if(l>=a&&r<=b){nd[root]+=c;sum[root]+=(len*c);return ;}pushdown(len,root);int mid=(l+r)/2;if(a<=mid)update(l,mid,root*2);if(b>mid)update(mid+1,r,root*2+1);sum[root]=sum[root*2]+sum[root*2+1];}int main(void){memset(nd,0,sizeof(nd));scanf("%lld%lld",&n,&m);build(1,n,1);for(ll i=0;i<m;i++){scanf("%s",s);if(s[0]=='Q'){scanf("%lld%lld",&a,&b);printf("%lld\n",query(1,n,1));}else{scanf("%lld%lld%lld",&a,&b,&c);update(1,n,1);}}}
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