HDU 4901 the romantic hero (DP)

Link: HDU 4901 the romantic hero

Given a sequence, two sets, S and T must be selected, and the subscript of the element selected in S must be smaller than the subscript of the element in T. In addition, the sum of elements in S must be equal to the sum of elements in T.

Problem-solving ideas: DP, the idea is very good, that is, the details are disgusting.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef __int64 ll;const int maxn = 1024;const ll MOD = 1e9+7;int n, num[maxn];ll l[maxn][maxn+10][2], r[maxn][maxn+10];void init () {    scanf("%d", &n);    for (int i = 1; i <= n; i++)        scanf("%d", &num[i]);}ll solve () {    memset(l, 0, sizeof(l));    memset(r, 0, sizeof(r));    for (int i = 1; i < n; i++) {        l[i][num[i]][1]++;        for (int j = 0; j < maxn; j++) {            l[i+1][j][1] = (l[i][j^num[i+1]][1] + l[i][j^num[i+1]][0])  % MOD;            l[i+1][j][0] = (l[i][j][1] + l[i][j][0]) % MOD;        }    }    for (int i = n; i; i--) {        r[i][num[i]]++;        for (int j = 0; j < maxn; j++) {            r[i-1][j] = (r[i-1][j] + r[i][j]) % MOD;            r[i-1][j&num[i-1]] = (r[i-1][j&num[i-1]] + r[i][j]) % MOD;        }    }    ll ans = 0;    for (int i = 1; i < n; i++) {        for (int j = 0; j < maxn; j++) {            /*            if (l[i][j] * r[i+1][j])                printf("%d %d %lld %lld\n", i, j, l[i][j], r[i+1][j]);                */            ans = (ans + l[i][j][1] * r[i+1][j] % MOD) % MOD;        }    }    return ans;}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        init();        printf("%I64d\n", solve());    }    return 0;}