codeforces 295a Greg and Array

codeforces 295a Greg and Array

​​ 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#define N 110000inline 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;}

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