CodeForces 408E Curious Array(组合数学+差分)
You've got an array consisting of n integers: a[1], a[2], ..., a[n]. Moreover, there are m queries, each query can be described by three integers li, ri, ki. Query li, ri, ki means that we should add to each element a[j], where li ≤ j ≤ ri.
Record means the binomial coefficient, or the number of combinations from yelements into groups of x elements.
You need to fulfil consecutively all queries and then print the final array.
Input
The first line contains integers n, m (1 ≤ n, m ≤ 105).
The second line contains n integers a[1], a[2], ..., a[n] (0 ≤ ai ≤ 109) — the initial array.
Next m lines contain queries in the format li, ri, ki — to all elements of the segment li... ri add number (1 ≤ li ≤ ri ≤ n; 0 ≤ k ≤ 100).
Output
Print n integers: the i-th number is the value of element a[i] after all the queries. As the values can be rather large, print them modulo 1000000007 (109 + 7).
Examples
5 1
0 0 0 0 0
1 5 0
1 1 1 1 1
10 2
1 2 3 4 5 0 0 0 0 0
1 6 1
6 10 2
2 4 6 8 10 7 3 6 10 15 题意:给出n个数,m次操作
每次操作给出li,ri,ki
将li-ri的范围里的第j个数加上 题解:考虑到上面的ki是在每个操作里是不会变的
考虑在杨辉三角中找规律
k=i是k=i-1的前缀和
然后可以考虑差分
在高维进行好差分后一维一维的降下来
注意差分的时候要每一层都差分 代码如下:
#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define mod 1000000007
using namespace std; long long fac[],inv[],n,m,a[][],sum[][]; long long kasumi(long long a,long long b)
{
long long ans=1ll;
while(b)
{
if(b&)
{
ans=ans*a%mod;
}
a=a*a%mod;
b>>=;
}
return ans;
} long long c(long long a,long long b)
{
return fac[a]*inv[b]%mod*inv[a-b]%mod;
} int main()
{
fac[]=,inv[]=;
for(int i=; i<=; i++)
{
fac[i]=fac[i-]*i*1ll%mod;
}
inv[]=kasumi(fac[],mod-);
for(int i=; i>=; i--)
{
inv[i]=inv[i+]*(i+)*1ll%mod;
}
scanf("%lld%lld",&n,&m);
for(int i=; i<n; i++)
{
scanf("%lld",&a[][i]);
}
long long l,r,k;
while(m--)
{
scanf("%lld%lld%lld",&l,&r,&k);
l--,k++;
a[k][l]++;
for(int i=; i<=k; i++)
{
a[i][r]-=c(k-i+r-l-,k-i);
a[i][r]=(a[i][r]+mod)%mod;
}
}
for(int k=; k>=; k--)
{
for(int i=;i<n;i++)
{
a[k][i]+=sum[k+][i+];
a[k][i]%=mod;
}
for(int i=;i<n;i++)
{
sum[k][i+]+=sum[k][i]+a[k][i];
sum[k][i+]%=mod;
}
}
for(int i=;i<n;i++)
{
printf("%lld ",a[][i]%mod);
}
}
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