codeforces 295a Greg and Array

http://www.elijahqi.win/archives/1189
Greg has an array a = a1, a2, …, an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, …, ri by value di.

Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, …, yi to the array.

Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.

Input
The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, …, an (0 ≤ ai ≤ 105) — the initial array.

Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).

Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).

The numbers in the lines are separated by single spaces.

Output
On a single line print n integers a1, a2, …, an — the array after executing all the queries. Separate the printed numbers by spaces.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

Examples
Input
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
Output
9 18 17
Input
1 1 1
1
1 1 1
1 1
Output
2
Input
4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
Output
5 18 31 20
做两个差分 就好 注意下 因为 我们最后要针对数值进行前缀和累加，所以我们应该一开始就处理出

他们的前缀和 第二次差分的时候应该在前缀和基础上差分

#include<cstdio>
#define N 110000
inline char gc(){
    static char now[1<<16],*S,*T;
    if (S==T){T=(S=now)+fread(now,1,1<<16,stdin);if (S==T) return EOF;}
    return *S++;
}
inline int read(){
    int x=0;char ch=gc();
    while (ch<'0'||ch>'9') ch=gc();
    while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();}
    return x;
}
long long a[N],sum[N];
int n,m,k,s[N],l[N],r[N],x[N];
int main(){
    freopen("cf.in","r",stdin);
    n=read();m=read();k=read();
    for (int i=1;i<=n;++i) a[i]=read();for (int i=1;i<=n;++i) sum[i]=a[i]-a[i-1];
    for (int i=1;i<=m;++i) l[i]=read(),r[i]=read(),x[i]=read();
    for (int i=1;i<=k;++i){
        int l1=read(),r1=read();s[l1]++;s[r1+1]--;
    }
    for (int i=1;i<=m;++i) s[i]+=s[i-1];
    for (int i=1;i<=m;++i) {
        sum[l[i]]+=(long long) x[i]*s[i];sum[r[i]+1]-=(long long) x[i]*s[i];
    } 
    for (int i=1;i<=n;++i) sum[i]+=sum[i-1];
    for (int i=1;i<=n;++i) printf("%I64d ",sum[i]); 
    return 0;
}