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1、HDU 5753 Permutation Bo

刚开始连题意没搞懂然后样例都搞不出来，搞懂题意后暴力找到规律推出来个公式，但是时间过去好久了，菜的抠脚啊。。。

题解的做法好厉害，%%%

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>
#include <bitset>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<int, int> Pair;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 1e3 + 10;
int n;
double c[maxn];

int main()
{
#ifdef __AiR_H
freopen("in.txt", "r", stdin);
#endif // __AiR_H
while (scanf("%d", &n) != EOF) {
double sum = 0;
for (int i = 1; i <= n; ++i) {
scanf("%lf", &c[i]);
sum += c[i];
}
if (n == 1) {
printf("%f\n", c[1]);
} else {
double ans = ((c[1] + c
) * 3.0 + (sum - c[1] - c
) * 2.0) / 6.0;
printf("%f\n", ans);
}
}
return 0;
}

2、HDU 5762 Teacher Bo

这道题怎么做都能过。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>
#include <bitset>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<int, int> Pair;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 1e5 + 10;

struct Node {
int x, y;
};

Node node[maxn];
int N, M;
map<int, int> Map;

int distance(int x1, int y1, int x2, int y2);

int main()
{
#ifdef __AiR_H
freopen("in.txt", "r", stdin);
#endif // __AiR_H
int T;
scanf("%d", &T);
while (T--) {
Map.clear();
scanf("%d%d", &N, &M);
for (int i = 0; i < N; ++i) {
scanf("%d%d", &node[i].x, &node[i].y);
}
bool flag = false;
if ((N*(N-1)/2) >= 2*M-1) {
flag = true;
} else {
for (int i = 0; i < N; ++i) {
for (int j = i+1; j < N; ++j) {
int t = distance(node[i].x, node[i].y, node[j].x, node[j].y);
if (Map[t] == 0) {
Map[t] = 1;
} else {
flag = true;
break;
}
}
if (flag) {
break;
}
}
}
if (flag) {
printf("YES\n");
} else {
printf("NO\n");
}
}
return 0;
}

int distance(int x1, int y1, int x2, int y2)
{
return abs(x1-x2) + abs(y1-y2);
}

3、HDU 5805 NanoApe Loves Sequence

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>
#include <bitset>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<int, int> Pair;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 1e5 + 10;
int A[maxn];
int n;

int main()
{
#ifdef __AiR_H
freopen("in.txt", "r", stdin);
#endif // __AiR_H
int T;
scanf("%d", &T);
while (T--) {
scanf("%d", &n);
int Max1 = -1, pos1 = 0, pos2 = 0, pos3 = 0, Max2 = -1, Max3 = -1;
for (int i = 0; i < n; ++i) {
scanf("%d", &A[i]);
if (i > 0) {
int t = abs(A[i] - A[i-1]);
if (t > Max1) {
Max3 = Max2;
pos3 = pos2;
Max2 = Max1;
pos2 = pos1;
Max1 = t;
pos1 = i;
} else if (t > Max2) {
Max3 = Max2;
pos3 = pos2;
Max2 = t;
pos2 = i;
} else if (t > Max3) {
Max3 = t;
pos3 = i;
}
}
}
ull ans = 0;
if (pos1 == 1) {
ans += Max2;
} else {
ans += Max1;
}
for (int i = 1; i < n-1; ++i) {
int t = abs(A[i+1] - A[i-1]);
if (pos1 == i && pos2 == i+1) {
ans += max(Max3, t);
} else if (pos2 == i && pos1 == i+1) {
ans += max(Max3, t);
} else if ((pos1 == i && pos3 == i+1) || (pos1 == i+1 && pos3 == i)) {
ans += max(Max2, t);
} else if ((pos2 == i && pos3 == i+1) || (pos2 == i+1 && pos3 == i)) {
ans += max(Max1, t);
} else if (pos1 == i || pos1 == i+1) {
ans += max(Max2, t);
} else {
ans += max(Max1, t);
}
}
if (pos1 == n) {
ans += Max2;
} else {
ans += Max1;
}
printf("%I64d\n", ans);
}
return 0;
}

4、HDU 5776 sum

 Provider:
HallStattMia

预处理前缀和，一旦有两个数模m的值相同，说明中间一部分连续子列可以组成m的倍数

另外，利用抽屉原理，我们可以得到，一旦n大于等于m，答案一定是YES

所以根据抽屉原理，这道题很好过

我是用了尺取法来做

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>
#include <bitset>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<int, int> Pair;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 1e5 + 10;
int num[maxn];

int main()
{
#ifdef __AiR_H
freopen("in.txt", "r", stdin);
#endif // __AiR_H
int T;
while (scanf("%d", &T) != EOF) {
while (T--) {
int n, m;
scanf("%d%d", &n, &m);
bool flag = false;
for (int i = 0; i < n; ++i) {
scanf("%d", &num[i]);
num[i] %= m;
if (num[i] == 0) {
flag = true;
}
}
queue<int> Q;
int sum = 0;
if (!flag) {
for (int i = 0; i < n; ++i) {
Q.push(num[i]);
sum += num[i];
if (sum%m == 0) {
flag = true;
break;
}
while (sum > m) {
sum -= Q.front();
Q.pop();
}
if (sum%m == 0) {
flag = true;
break;
}
}
}
if (flag) {
printf("YES\n");
} else {
printf("NO\n");
}
}
}
return 0;
}

5、UVa 11292 Dragon of Loowater

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>
#include <bitset>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<int, int> Pair;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
int n, m;
vector<int> V1, V2;

int main()
{
#ifdef __AiR_H
freopen("in.txt", "r", stdin);
#endif // __AiR_H
while (scanf("%d%d", &n, &m) != EOF && !(n ==0 && m == 0)) {
V1.clear(), V2.clear();
int num;
while (n--) {
scanf("%d", &num);
V1.push_back(num);
}
while (m--) {
scanf("%d", &num);
V2.push_back(num);
}
if (n > m) {
printf("Loowater is doomed!\n");
} else {
sort(V1.begin(), V1.end()), sort(V2.begin(), V2.end());
ull ans = 0;
int Size1 = V1.size(), Size2 = V2.size();
int pos1 = 0, pos2 = 0;
while (pos1 < Size1 && pos2 < Size2) {
if (V2[pos2] >= V1[pos1]) {
ans += V2[pos2];
++pos1;
++pos2;
} else {
++pos2;
}
}
if (pos1 == Size1) {
printf("%llu\n", ans);
} else {
printf("Loowater is doomed!\n");
}
}
}
return 0;
}

6、UVa 11729  Commando War

假设我们交换两个相邻的任务 X 和 Y（交换前 X 在 Y 之前，交换后 Y 在 X 之前），不难发现其他任务的完成时间没有影响，那么这两个任务呢？

情况一：交换之前，任务 Y 比 X 先结束，如图（a）所示。不难发现，交换之后 X的结束时间延后，Y 的结束时间提前，最终答案不会变好

情况二：交换之前，X 比 Y 先结束，因此交换后答案变好的充要条件是：交换后 X 的结束时间比交换前 Y 的结束时间早（交换后 Y 的结束时间肯定变早了）

如图（b）所示。这个条件可以写成 B[Y]+B[X]+J[X] < B[X]+B[Y]+J[Y]，化简得 J[X] < J[Y] ，这就是我们贪心的依据

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>
#include <bitset>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<int, int> Pair;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 1e3 + 10;

struct Node {
int B, J;
};

Node node[maxn];
int N;

bool cmp(Node a, Node b);

int main()
{
#ifdef __AiR_H
freopen("in.txt", "r", stdin);
#endif // __AiR_H
int Case = 0;
while (scanf("%d", &N) != EOF && N != 0) {
for (int i = 0; i < N; ++i) {
scanf("%d%d", &node[i].B, &node[i].J);
}
sort(node, node+N, cmp);
ull ans = 0;
ull B_sum = 0;
for (int i = 0; i < N; ++i) {
B_sum += node[i].B;
ull ans_t = B_sum + node[i].J;
if (ans_t > ans) {
ans = ans_t;
}
}
printf("Case %d: %llu\n", ++Case, ans);
}
return 0;
}

bool cmp(Node a, Node b)
{
return a.J > b.J;
}

7、Codeforces Gym 100860A  AutoCoder

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <cmath>
#include <cctype>
#include <bitset>
#include <ctime>

using namespace std;

#define REP(i, n) for (int i = 0; i < (n); ++i)

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<int, int> Pair;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;

int main()
{
#ifdef __AiR_H
freopen("in.txt", "r", stdin);
#endif // __AiR_H
ll n;
scanf("%I64d", &n);
if (n == 0) {
printf("0\n");
} else {
--n;
ll ans = 0, t = 0;
while (n) {
if (n <= 2) {
ans += 1;
break;
}
if (n%2 == 0) {
t = n / 2;
ans += t * (t+1) / 2;
n = (n - 2) / 2;
} else {
t = (n+1) / 2;
ans += t * (t+1) / 2;
n = (n-3) / 2;
}
}
printf("%I64d\n", ans);
}
return 0;
}