hust 1422 Candy!

题目描述

Halloween
is coming! So xiaoY has to prepare M candies to treat the neighbor
kids.

When
this horrible night come, There are N little children ask xiaoY for candy, and
each of them has different demand. For the i-th kid, his (or her) candy must not
less than Min[i], and not greater than Max[i].

Now
xiaoY wants to know how many ways he can give out all of his candies. (in case
the large answer, please tell xiaoY the answer modular
10000007)

输入

First
line is an integer T, represents T cases followed:

N,
M ---- Number of children, Number of candies.

(0<N<=20,
0<M<=10000)

Min[i],
Max[i] ---- The lower bound and upper bound of I-th kid’s demand. (N
lines)

(0<=Min[i]
<= Max[i] <=M)

输出

Output
a number W which means total different ways of handing out the candies. (%
10000007)

样例输入

2

3 5

1 2

1 2

1 2

3 6

1 2

1 2

1 2

样例输出

Case #1: 3

Case #2: 1

神一样的题目，当然是dp解决了，dp[i+j]+=dp[j],表示将i放在当前的人，j给i前面的人，很好理解吧！我9000多ms过的，当然还有16ms过的，膜拜

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

const int maxn=10000+10;

const int mod=10000007;

int f1[maxn],f2[maxn];
int main()

{

    int t,n,m,a,b,MAX;

    scanf("%d",&t);

    for (int tt=1;tt<=t;tt++)

    {

        memset(f1,0,sizeof(f1));

        memset(f2,0,sizeof(f2));

        scanf("%d%d",&n,&m);

        scanf("%d%d",&a,&b);

        for (int i=a;i<=b;i++) f1[i]=1;

        MAX=b;

        for (int k=2;k<=n;k++)

        {

            scanf("%d%d",&a,&b);

            for (int i=0;i<=MAX && i<=m;i++)

            {

                for (int j=a;i+j<=m && j<=b;j++)

                {

                    f2[i+j]+=f1[i];

                    f2[i+j]=f2[i+j]%mod;

                }

            }

            MAX+=b;

            for (int i=0;i<=MAX && i<=m;i++)

            {

                f1[i]=f2[i];

                f2[i]=0;

            }

        }

        printf("Case #%d: %d\n",tt,f1[m]%mod);

    }

    return 0;

}

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