标签：com cst inpu second build interval efi sim nes
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
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #define lson l , m , rt << 1 6 #define rson m+1,r , rt << 1 | 1 7 using namespace std; 8 typedef long long ll; 9 10 const int maxn=200000; 11 12 int n,q; 13 ll sum[maxn<<2],add[maxn<<2]; 14 15 void Pushup(int rt){ sum[rt]=sum[rt<<1]+sum[(rt<<1)|1]; } 16 17 void Pushdown(int rt,int m){ 18 if(add[rt]){ 19 add[rt<<1]+=add[rt]; 20 sum[rt<<1]+=(m-(m>>1))*add[rt]; 21 add[(rt<<1)|1]+=add[rt]; 22 sum[(rt<<1)|1]+=(m>>1)*add[rt]; 23 add[rt]=0; 24 } 25 } 26 27 void Build(int l,int r,int rt){ 28 add[rt]=0; 29 if(l==r) { scanf("%lld",&sum[rt]); return; } 30 int m=(l+r)>>1; 31 Build(lson); 32 Build(rson); 33 Pushup(rt); 34 } 35 36 void Update(int L,int R,int c,int l,int r,int rt){ 37 if(L<=l&&r<=R){ 38 add[rt]+=c; 39 sum[rt]+=(ll)c*(r-l+1); 40 return ; 41 } 42 Pushdown(rt,r-l+1); 43 int m=(l+r)>>1; 44 if(L<=m) Update(L,R,c,lson); 45 if(R>m) Update(L,R,c,rson); 46 Pushup(rt); 47 } 48 49 ll Query(int L,int R,int l,int r,int rt){ 50 if(L<=l&&r<=R) return sum[rt]; 51 Pushdown(rt,r-l+1); 52 int m=(l+r)>>1; 53 ll temp=0; 54 if(L<=m) temp+=Query(L,R,lson); 55 if(R>m) temp+=Query(L,R,rson); 56 return temp; 57 } 58 59 int main() 60 { cin>>n>>q; 61 Build(1,n,1); 62 while(q--){ 63 char op[2]; 64 scanf("%s",op); 65 int a,b,c; 66 if(op[0]==‘Q‘){ 67 scanf("%d%d",&a,&b); 68 printf("%lld\n",Query(a,b,1,n,1)); 69 } 70 if(op[0]==‘C‘){ 71 scanf("%d%d%d",&a,&b,&c); 72 Update(a,b,c,1,n,1); 73 } 74 } 75 return 0; 76 }
A Simple Problem with Integers POJ - 3468
标签：com cst inpu second build interval efi sim nes
原文地址：http://www.cnblogs.com/zgglj-com/p/7364913.html
