以后每做完一场CF，解题报告都写在一起吧
暴力||二分 A - Bear and Elections
题意：有n个候选人，第一个候选人可以贿赂其他人拿到他们的票，问最少要贿赂多少张票第一个人才能赢
分析：正解竟然是暴力！没敢写暴力，卡了很久，导致这场比赛差点爆零！二分的话可以优化，但对于这题来说好像不需要。。。
收获：以后CF div2的A题果断暴力
代码(暴力)：
/************************************************
* Author :Running_Time
* Created Time :2015-8-30 0:40:46
* File Name :A.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N];
int n, x;
int main(void) {
scanf ("%d", &n);
scanf ("%d", &a[0]); n--;
for (int i=1; i<=n; ++i) scanf ("%d", &a[i]);
int ans = 0;
while (true) {
int mxi = 0;
for (int i=1; i<=n; ++i) {
if (a[i] >= a[mxi]) mxi = i;
}
if (mxi == 0) break;
ans++; a[0]++; a[mxi]--;
}
printf ("%d
", ans);
return 0;
}
代码(二分):
/************************************************
* Author :Running_Time
* Created Time :2015-8-30 0:40:46
* File Name :A.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N];
int n;
bool cmp(int x, int y) {
return x > y;
}
bool check(int add) {
int y = a[0] + add;
for (int i=1; i<=n; ++i) {
if (a[i] >= y) {
if (add <= 0) return false;
add -= (a[i] - (y - 1));
if (add < 0) return false;
}
else break;
}
return true;
}
int main(void) {
scanf ("%d", &n);
scanf ("%d", &a[0]); n--;
for (int i=1; i<=n; ++i) scanf ("%d", &a[i]);
sort (a+1, a+1+n, cmp);
if (a[0] > a[1]) {
puts ("0"); return 0;
}
int l = 0, r = 1000;
while (l <= r) {
int mid = (l + r) >> 1;
if (check (mid)) r = mid - 1;
else l = mid + 1;
}
printf ("%d
", l);
return 0;
}
暴力 B - Bear and Three Musketeers
题意：找一个三元环并且三个点的度数和-6最小
分析：三元环做过一题。这题能暴力跑过，DFS不用写。枚举边的两个端点，再找是否存在另外一个点构成三元环就可以了
收获：CF div2 B也暴力过
代码：
/************************************************
* Author :Running_Time
* Created Time :2015-8-25 19:24:24
* File Name :E_topo.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
typedef pair<int, int> PII;
const int N = 4e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int deg[N];
bool g[N][N];
vector<PII> G;
int n, m;
int main(void) {
scanf ("%d%d", &n, &m);
G.clear ();
memset (g, false, sizeof (g));
memset (deg, 0, sizeof (deg));
for (int u, v, i=1; i<=m; ++i) {
scanf ("%d%d", &u, &v);
G.push_back (PII (u, v));
g[u][v] = true; g[v][u] = true;
deg[u]++; deg[v]++;
}
int ans = INF;
for (int i=0; i<G.size (); ++i) {
int u = G[i].first, v = G[i].second;
for (int j=1; j<=n; ++j) {
if (g[u][j] && g[v][j]) {
ans = min (ans, deg[u] + deg[v] + deg[j] - 6);
}
}
}
printf ("%d
", (ans == INF) ? -1 : ans);
return 0;
}
题意：给n个数字，每个数字能*2或*3(可多次)，问是否能使得n个数相等
分析：要达到相等，那么每个数字除了2和3的数字外的剩余数字要相等，一个数除2或3是log级别的? 复杂度不会分析。。。
收获：其实这题很水，想到/2 /3就行了
代码：
/************************************************
* Author :Running_Time
* Created Time :2015-8-30 1:42:08
* File Name :C.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N];
int GCD(int a, int b) {
return b ? GCD (b, a % b) : a;
}
int main(void) {
int n; scanf ("%d", &n);
for (int i=1; i<=n; ++i) scanf ("%d", &a[i]);
bool flag = true;
for (int i=1; i<=n; ++i) {
while (a[i] % 2 == 0) a[i] /= 2;
while (a[i] % 3 == 0) a[i] /= 3;
}
for (int i=1; i<n; ++i) {
if (a[i] != a[i+1]) {
flag = false; break;
}
}
puts (flag ? "Yes" : "No");
return 0;
}
题意：有n列高度不等的方块，每次可以将和空气(至少一面不和砖头或地面接触)接触的搬走，问要搬几次
分析：先考虑一列方块时只要搬一次。两列时，如果第二列高度大于第一列，那么搬两次
，否则只搬一次。三列的情况同理
问题转换成最长连续上升序列，从前往后扫一次还有从后往前扫一次，两次取最小值，答案取最大值
代码：
/************************************************
* Author :Running_Time
* Created Time :2015-8-30 16:56:38
* File Name :D.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N], dp1[N], dp2[N];
int main(void) {
int n; scanf ("%d", &n);
for (int i=0; i<n; ++i) scanf ("%d", &a[i]);
dp1[0] = 1;
for (int i=1; i<n; ++i) dp1[i] = min (a[i], dp1[i-1] + 1);
dp2[n-1] = 1;
for (int i=n-2; i>=0; --i) dp2[i] = min (a[i], dp2[i+1] + 1);
int ans = 0;
for (int i=0; i<n; ++i) {
ans = max (ans, min (dp1[i], dp2[i]));
}
printf ("%d
", ans);
return 0;
}