WHU 1579 Big data (DP)

题意：

f[0]=0,f[i]=f[i-1]+a or b.

求满足L<=∑f[n]<=R的序列的种数

n<100.  |a|,|b|<=10000.  |L|,|R|<1e9

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Solution

其实就是一个背包问题.

当a=b,序列 0 a a+a和 0 b b+b 竟然算不同的序列= =,巨坑

#include <iostream>
#define LL long long
using namespace std;
const int MOD = (int) 1e9 + ;
const int M = ;

LL N, a, b, L, R;
LL dp[][ * ];

int main() {
    dp[][] = ;
    for (int i = ; i <= M; i++)
        for (int j = ; j <= (i * (i - ) / ); j++) {
            dp[i][j] = (dp[i - ][j] + dp[i][j]) % MOD;
            dp[i][j + i] = (dp[i - ][j] + dp[i][j + i]) % MOD;
        }

    for (int i = ; i <= M; i++)
        for (int j =  (i + ) * i /  - ; j >= ; j--)
            dp[i][j] = (dp[i][j + ] + dp[i][j]) % MOD;

    while (cin >> N >> a >> b >> L >> R) {
        if (a > b) swap (a, b);
        LL s = a * ( + N) * N / ;
        if (s <= R && b != a ) {
            LL nl = (L - s ) / (b - a), nr = (R - s) / (b - a) + ;
            if ( (L - s) % (b - a) != ) nl++;
            if (L - s <= ) nl = ;
            if (nl > (N + ) *N / ) nl = (N + ) * N /   + ;
            if (nr > (N + ) *N / ) nr = (N + ) * N /  + ;
            cout << (MOD + dp[N][nl] - dp[N][nr]) % MOD << endl;
        }
        else if (s >= L && s <= R && b == a) {
            LL ans = ;
            for (int i = ; i <= N; i++)
                ans = (ans * ) % MOD;
            cout << ans << endl;
        }
        else
            cout <<  << endl;
    }
    return ;
}