whu oj 1579 - Big data

Description

The data scientists of CVTE declared that they had discovered a new number sequence system. The numbers sequences can be described as follows: they all have N numbers and their deferred expenses are A(i) = A(i-1) + x, where x equals a or b for different i. ( A(i) = 0 )

Now we are wondering that how many sequences satisfy that sigma( Ai ) 1 <= i <= n is between L and R.

Please module your final answer by 10^9 + 7.

Input

No more than 20000 test cases.

The first line of each test case contains five number N, a, b, L and R. ( 2 <= N <= 100, -10000 <= a, b <= 10000, -10^9 <= L <= R <= 10^9 )

Output

Output your answer on a single line for each case.

Sample Input

5 1 2 0 15

Sample Output

这道题目出的水平还是有的，当时武大校赛时没做出来，坑了队友，果然不训练真的是不行的。

题源：http://acm.whu.edu.cn/land/problem/detail?problem_id=1579

题意：

A(i) = A(i-1) + a or b，求满足L <= ∑(Ai) <= R的有多少种。

思路：

相当于求(a or b) * n + (a or b) * (n - 1)…..(a or b) <= R
step 1: 令c = b - a，那么相当于(c or 0) * n + (c or 0) * (n - 1) + ….(c or 0) <= R - a* n * (n + 1) / 2
step 2：令X = R - a* n * (n + 1) / 2，相当于(1 or 0)* n + (1 or 0) * (n - 1)….(1 or 0) <= X / c
step 3：剩下一个简单dp计数问题，预处理即可。

至于a=b时，需要特判，我觉得序列都是一样的，但是出题人好像作为不同处理的 = =

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define mod 1000000007
typedef long long ll;
using namespace std;

ll n,m;
ll a,b,le,ri;
ll dp[105][5055],sum[105][5055];

ll getdp(ll x)
{
    if(x<0) return 0;
    if(x>=5050) x=5050;
    return sum[n][x];
}
int main()
{
    ll i,j;
    dp[0][0]=1;
    for(i=0;i<100;i++)
    {
        for(j=0;j<=i*(i+1)/2;j++)
        {
            if(dp[i][j]==0) continue ;
            dp[i+1][j]+=dp[i][j];
            dp[i+1][j]%=mod;
            dp[i+1][j+(i+1)]+=dp[i][j];
            dp[i+1][j+(i+1)]%=mod;
        }
    }
    for(i=0;i<=100;i++)
    {
        sum[i][0]=dp[i][0];
        for(j=1;j<=5050;j++)
        {
            sum[i][j]=sum[i][j-1]+dp[i][j];
            sum[i][j]%=mod;
        }
    }
    while(~scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&le,&ri))
    {
        if(a==b)
        {
            if((n+1)*n/2*a>=le&&(n+1)*n/2*a<=ri) printf("%lld\n",sum[n][n*(n+1)/2]);
            else printf("0\n");
            continue ;
        }
        if(a>b) swap(a,b);
        ll c=b-a;
        ll y;
        if((ri-a*n*(n+1)/2)<0) y=-1;
        else y=(ri-a*n*(n+1)/2)/c;
        ll x;
        if((le-a*n*(n+1)/2)<0) x=-1;
        else
        {
            if((le-a*n*(n+1)/2)%c==0) x=(le-a*n*(n+1)/2)/c-1;
            else x=(le-a*n*(n+1)/2)/c;
        }
        ll ans=getdp(y)-getdp(x);
        ans=(ans+mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}
/*
3 1 3 -7 6
3 1 3 0 15
*/

原文地址：http://blog.csdn.net/tobewhatyouwanttobe/article/details/45933097