A
第一题明显统计，注意0和long long（我WA，RE好几次）
/*
* Problem: A. Matrix
* Author: Shun Yao
*/
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <assert.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <bitset>
#include <utility>
#include <iomanip>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
//using namespace std;
int a, n, b[4444], c[44444];
char s[4444];
int main(/*int argc, char **argv*/) {
int i, j, x;
long long ans;
scanf("%d", &a);
scanf(" %s", s + 1);
n = strlen(s + 1);
for (i = 1; i <= n; ++i)
b[i] = s[i] - '0';
memset(c, 0, sizeof c);
ans = 0;
for (i = 1; i <= n; ++i) {
x = 0;
for (j = i; j <= n; ++j) {
x += b[j];
++c[x];
}
}
if (a) {
for (i = 1; i <= 40000 && i * i <= a; ++i)
if (a % i == 0 && a / i <= 40000)
ans += static_cast<long long>(c[i]) * c[a / i] * (i * i == a ? 1 : 2);
} else
ans = static_cast<long long>(c[0]) * (n * (n + 1) - c[0]);
printf("%I64d", ans);
fclose(stdin);
fclose(stdout);
return 0;
}
B
用背包算出每个价值是否出现过，然后贪心即可。
/*
* Problem: B. Free Market
* Author: Shun Yao
*/
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <assert.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <bitset>
#include <utility>
#include <iomanip>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
//using namespace std;
int n, d, c[55], f[500010];
int main(/*int argc, char **argv*/) {
int i, j, sum, ans1, ans2;
scanf("%d%d", &n, &d);
sum = 0;
for (i = 1; i <= n; ++i) {
scanf("%d", c + i);
sum += c[i];
}
memset(f, 0, sizeof f);
f[0] = 1;
for (i = 1; i <= n; ++i)
for (j = sum; j >= c[i]; --j)
f[j] |= f[j - c[i]];
ans1 = 0;
ans2 = 0;
while (ans1 < sum) {
for (j = ans1 + d <= sum ? ans1 + d : sum; j > ans1; --j)
if (f[j])
break;
if (j <= ans1)
break;
ans1 = j;
++ans2;
}
printf("%d %d", ans1, ans2);
fclose(stdin);
fclose(stdout);
return 0;
}
C
不懂证明，求解释。。。
D
做法是随机算法（没有想到啊。）
/*
* Problem: D. Ghd
* Author: Shun Yao
*/
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <assert.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <bitset>
#include <utility>
#include <iomanip>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
//using namespace std;
const int MAXN = 1000010;
int n;
long long a[MAXN], ans;
std::vector<long long> e, v;
long long gcd(long long a, long long b) {
return !b ? a : gcd(b, a % b);
}
int main(/*int argc, char **argv*/) {
int i, j, x;
// freopen("D.in", "r", stdin);
// freopen("D.out", "w", stdout);
scanf("%d", &n);
for (i = 1; i <= n; ++i)
scanf("%I64d", a + i);
srand((unsigned)time(0));
ans = 0;
while (clock() < 3000L) {
x = static_cast<long long>(rand()) * rand() % n + 1;
e.clear();
v.clear();
for (i = 1; static_cast<long long>(i) * i <= a[x]; ++i)
if (a[x] % i == 0) {
e.push_back(i);
if (static_cast<long long>(i) * i != a[x])
e.push_back(a[x] / i);
}
std::sort(e.begin(), e.end());
v.resize(e.size(), 0);
for (i = 1; i <= n; ++i)
++v[std::lower_bound(e.begin(), e.end(), gcd(a[x], a[i])) - e.begin()];
for (i = 0; i < static_cast<int>(e.size()); ++i) {
for (j = i + 1; j < static_cast<int>(e.size()); ++j)
if (e[j] % e[i] == 0)
v[i] += v[j];
if (v[i] + v[i] >= n)
ans = std::max(ans, e[i]);
}
}
printf("%I64d", ans);
fclose(stdin);
fclose(stdout);
return 0;
}
E
二维分治。
/*
* Problem: E. Empty Rectangles
* Author: Shun Yao
*/
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <assert.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <bitset>
#include <utility>
#include <iomanip>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
//using namespace std;
int n, m, k;
char s[2505][2505];
long long a[2505][2505];
inline long long sum(int r1, int c1, int r2, int c2) {
return a[r2][c2] - a[r1 - 1][c2] - a[r2][c1 - 1] + a[r1 - 1][c1 - 1];
}
long long go(int r1, int c1, int r2, int c2) {
if (r1 == r2 && c1 == c2)
return sum(r1, c1, r2, c2) == k;
if (r2 - r1 > c2 - c1) {
long long cnt = 0;
int m = (r1 + r2) >> 1;
int up[k + 1], dw[k + 1];
for (int i = c1; i <= c2; i++) {
for (int x = 0; x <= k; x++)up[x] = m - r1 + 1, dw[x] = r2 - m;
for (int j = i; j <= c2; j++) {
for (int x = 0; x <= k; x++) {
while (up[x] && sum(m - up[x] + 1, i, m, j) > x)
--up[x];
while (dw[x] && sum(m + 1, i, m + dw[x], j) > x)
--dw[x];
}
for (int x = 0; x <= k; x++) {
cnt += (long long)(up[x] - (x ? up[x - 1] : 0)) * (dw[k - x] - (k - x ? dw[k - x - 1] : 0));
}
}
}
return cnt + go(r1, c1, m, c2) + go(m + 1, c1, r2, c2);
} else {
long long cnt = 0;
int m = (c1 + c2) >> 1;
int up[k + 1], dw[k + 1];
for (int i = r1; i <= r2; i++) {
for (int x = 0; x <= k; x++)up[x] = m - c1 + 1, dw[x] = c2 - m;
for (int j = i; j <= r2; j++) {
for (int x = 0; x <= k; x++) {
while (up[x] && sum(i, m - up[x] + 1, j, m) > x)up[x]--;
while (dw[x] && sum(i, m + 1, j, m + dw[x]) > x)dw[x]--;
}
for (int x = 0; x <= k; x++) {
cnt += (long long)(up[x] - (x ? up[x - 1] : 0)) * (dw[k - x] - (k - x ? dw[k - x - 1] : 0));
}
}
}
return cnt + go(r1, c1, r2, m) + go(r1, m + 1, r2, c2);
}
}
int main() {
freopen("E.in", "r", stdin);
freopen("E.out", "w", stdout);
scanf("%d%d%d", &n, &m, &k);
for (int i = 1; i <= n; i++)scanf("%s", &s[i][1]);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1] + s[i][j] - '0';
}
}
printf("%I64d", go(1, 1, n, m));
fclose(stdin);
fclose(stdout);
return 0;
}