hdu 5435 A serious math problem（数位dp）

题目链接：hdu 5435 A serious math problem

裸的数位dp。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 100005;
const int maxd = 16;
const int mod = 1e9 + 7;

ll dp[maxn][maxd + 5];
char a[maxn], b[maxn];

int get(char* s) {
	int n = strlen(s), ret = 0;
	for (int i = 0; i < n; i++)
		ret ^= (s[i]-'0');
	return ret;
}

ll solve (char* s) {
	int n = strlen(s), pre = 0;
	memset(dp, 0, sizeof(dp));

	for (int i = 0; i < n; i++) {
		if (i) {
			for (int x = 0; x < maxd; x++) {
				for (int y = 0; y < 10; y++) {
					int v = x^y;
					dp[i][v] = (dp[i][v] + dp[i-1][x]) % mod;
				}
			}
		}

		for (int x = 0; x < s[i]-'0'; x++) {
			int v = pre^x;
			dp[i][v] = (dp[i][v] + 1) % mod;
		}
		pre ^= s[i]-'0';
	}

	ll ret = 0;
	for (int i = 0; i < maxd; i++)
		ret = (ret + dp[n-1][i] * i) % mod;
	return ret;
}

int main () {

	int cas;
	scanf("%d", &cas);
	for (int kcas = 1; kcas <= cas; kcas++) {
		scanf("%s%s", a, b);
		ll l = solve(a);
		ll r = solve(b);
		ll ans = ((r - l + mod) % mod + get(b)) % mod;
		printf("Case #%d: %d\n", kcas, (int)ans);
	}
	return 0;
}