BZOJ-1042 [HAOI2008]硬币购物 容斥原理 +０１背包方案数

大家都很强，可与之共勉。

[HAOI2008]硬币购物

Description

　　硬币购物一共有4种硬币。面值分别为c1,c2,c3,c4。某人去商店买东西，去了tot次。每次带di枚ci硬币，买si的价值的东西。请问每次有多少种付款方法。

Input

　　第一行 c1,c2,c3,c4,tot 下面tot行 d1,d2,d3,d4,s,其中di,s<=100000,tot<=1000

Output

每次的方法数

Sample Input

1 2 5 10 2

3 2 3 1 10

1000 2 2 2 900

Sample Output

4

27

**************************************************************    Problem: 1042    User: Lazer2001    Language: C++    Result: Accepted    Time:40 ms    Memory:1604 kb****************************************************************/#include <cstdio>unsigned int c[4], d[4], S, T;unsigned long long dp[100005] = {1}, ans;void Dfs ( short step, short flag, int sum )  {    if ( sum < 0 )   return;    if ( step == 4 )  {        flag & 1 ?  ans -= dp[sum] : ans += dp[sum];        return;    }       Dfs ( step + 1, flag, sum );    Dfs ( step + 1, flag ^ 1, sum - ( d[step] + 1 ) * c[step] );}int main ( )  {    for ( int i = 0; i < 4; ++i )  scanf ( "%d", c + i );    for ( int i = 0; i < 4; ++i )        for ( int j = c[i]; j <= 100000; ++j )            dp[j] += dp[j - c[i]];    for ( scanf ( "%d", &T ); T; --T )  {        for ( int i = 0; i < 4; ++i )  scanf ( "%d", d + i );        scanf ( "%d", &S );        ans = 0;        Dfs ( 0, 0, S );        printf ( "%lld\n", ans );    }}