D - Sum of Large Numbers

Time Limit: 2 sec / Memory Limit: 1024 MB

 

Score : $400$ points

ps:我发现一个很有趣的问题，long long和int 一起使用时，数据过大可能会出现错误，比如下面。所以最好还是统一用long long。

Problem Statement

We have $\mathrm{N+1}$integers: ${\mathrm{10^100}}^{}$, ${\mathrm{10^100+1}}^{}$, ..., ${\mathrm{10^100+N}}^{}$.

We will choose $K$ or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo $(109+7)$.

Constraints

• $\mathrm{1\le N\le 2\times 105}$
• $\mathrm{1\le K\le N+1}$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$$K$

Output

Print the number of possible values of the sum, modulo $(10^9+7)$.

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Sample Input 1 Copy

Copy

3 2

Sample Output 1 Copy

Copy

10

The sum can take $1010$ values, as follows:

• $(10100)+(10100+1)=2\times 10100+1(10100)+(10100+1)=2\times 10100+1$
• $(10100)+(10100+2)=2\times 10100+2(10100)+(10100+2)=2\times 10100+2$
• $(10100)+(10100+3)=(10100+1)+(10100+2)=2\times 10100+3(10100)+(10100+3)=(10100+1)+(10100+2)=2\times 10100+3$
• $(10100+1)+(10100+3)=2\times 10100+4(10100+1)+(10100+3)=2\times 10100+4$
• $(10100+2)+(10100+3)=2\times 10100+5(10100+2)+(10100+3)=2\times 10100+5$
• $(10100)+(10100+1)+(10100+2)=3\times 10100+3(10100)+(10100+1)+(10100+2)=3\times 10100+3$
• $(10100)+(10100+1)+(10100+3)=3\times 10100+4(10100)+(10100+1)+(10100+3)=3\times 10100+4$
• $(10100)+(10100+2)+(10100+3)=3\times 10100+5(10100)+(10100+2)+(10100+3)=3\times 10100+5$
• $(10100+1)+(10100+2)+(10100+3)=3\times 10100+6(10100+1)+(10100+2)+(10100+3)=3\times 10100+6$
• $(10100)+(10100+1)+(10100+2)+(10100+3)=4\times 10100+6(10100)+(10100+1)+(10100+2)+(10100+3)=4\times 10100+6$

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Sample Input 2 Copy

Copy

200000 200001

Sample Output 2 Copy

Copy

1

We must choose all of the integers, so the sum can take just $1$value.

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Sample Input 3 Copy

Copy

141421 35623

Sample Output 3 Copy

Copy

220280457

#include <bits/stdc++.h>
using namespace std;
const int mod=1e9+7;

int main(){
long long n,k;//若使用int n,k,答案错误
long long sum=0;
cin>>n>>k;
while(k<=n+1){
    sum+=(2*n-k+1)*k/2-k*(k-1)/2+1;
    k++;
}
sum%=mod;
cout<<sum<<endl;
return 0;
}