[FOJ][2042][The Mad Mathematician][数位dp]

/*
题目：The Mad Mathematician
题目来源：FOJ 2042
题目难度：难
题目内容或思路：
 数位dp + xor

 已知A，B，C，D，E五个数(long long型)
 求下面程序所得sum的值
 for (int i = A; i <= B; ++i)
 for (int j = C; j <= D; ++j)
 if ((i ^ j) > E)
 sum += i ^ j;
 
 dp[i][j] 表示第i位，状态为j时的情况数
 sum[i][j] 表示第i位，状态为j时的sum值
 第二维的状态分别用3个bit表示 
 第一个数i位之前是否已经小于他
 第二个数i位之前是否已经小于他
 E的i位之前是否已经大于他
做题日期：2011.6.4
*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <ctime>
#include <cstdlib>
#include <bitset>
#include <algorithm>
using namespace std;

typedef long long ll;
#define R "%I64d"
const int mod = 1000000007;
ll dp[64][8], sum[64][8];

ll gao(ll a, ll b, ll e) {
    if (a < 0 || b < 0) return 0;
    memset(dp, 0, sizeof(dp));
    memset(sum, 0, sizeof(sum));
    dp[63][0] = 1;
    for (int i = 62; i >= 0; --i) {
        ll z = 1LL << i;
        for (int pre = 0; pre < 8; ++pre) {
            ll cnt = dp[i + 1][pre];
            ll val = sum[i + 1][pre];
            if (cnt == 0) continue;
            for (int st = 0; st < 4; ++st) {
                int u = st & 1, v = st >> 1;
                int cur = 0;

                if (pre & 2) cur |= 2;
                else if (u == 0 && (a & z)) cur |= 2;
                else if (u && (a & z) == 0) continue;

                if (pre & 4) cur |= 4;
                else if (v == 0 && (b & z)) cur |= 4;
                else if (v && (b & z) == 0) continue;

                if (pre & 1) cur |= 1;
                else if ((u ^ v) && (e & z) == 0) cur |= 1;
                else if ((u ^ v) == 0 && (e & z)) continue;

                dp[i][cur] = (dp[i][cur] + cnt) % mod;
                if (u ^ v)
                    sum[i][cur] = (sum[i][cur] + val + z % mod * cnt) % mod;
                else
                    sum[i][cur] = (sum[i][cur] + val) % mod;
            }
        }
    }
    ll ans = 0;
    for (int i = 1; i < 8; i += 2)
        ans = (ans + sum[0][i]) % mod;
    return ans;
}

void solve() {
    ll a, b, c, d, e;
    scanf(R R R R R, &a, &b, &c, &d, &e);
    ll ans = gao(b, d, e) - gao(b, c - 1, e)
            - gao(a - 1, d, e) + gao(a - 1, c - 1, e);
    ans = (ans % mod + mod) % mod;
    printf(R "\n", ans);
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in", "r", stdin);
#endif
    int cas;
    scanf("%d", &cas);
    for (int i = 1; i <= cas; ++i) {
        printf("Case %d: ", i);
        solve();
    }
    return 0;
}