T1
思路
因为0的个数超过了一半，所以只要将拍完序后，最中间的数到想得到的中位数之间的每个数都变成S即可。
代码
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<ctime>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 100000 + 100;
ll read() {
ll x = 0, f = 1;
char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
ll ans;
ll a[N];
int main() {
freopen("median.in","r",stdin);
freopen("median.out","w",stdout);
int n = read();
ll s = read();
for(int i = 1;i <= n;++i) a[i] = read();
sort(a + 1,a + n + 1);
for(int i = n/2 + 1;i <= n;++i) {
if(a[i] >= s) break;
ans += s - a[i];
}
cout<<ans;
return 0;
}
T2
60分思路
对于前60分，可以直接(n^2)dp。用f[i][j]表示前i天，其中j天写了作业的方案数。这样没写作业的天数就是i-j,根据要求((i-j)-j<k => j >i-k)只要控制一下转移的边界就行了。
100分思路
对于一百分，$$C({2n}^n) - C({2n}^{n+k})$$
然后分解质因数求组合数。
60分代码
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<ctime>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1000000 + 100;
int mod;
ll read() {
ll x = 0, f = 1;
char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
int n, K;
namespace BF1 {
ll f[2010][2010];
void Main() {
f[0][0] = 1;
for(int i = 1;i <= n * 2;++i) {
if(i < K) f[i][0] = 1;
for(int j = max(((i - K)/2 + 1),1);j <= min(i,n);++j) {
f[i][j] = f[i - 1][j] + f[i - 1][j - 1];
f[i][j] >= mod ? f[i][j] -= mod : 0;
}
}
cout<<f[n * 2][n];
return;
}
}
int ff[N];
int main() {
freopen("term.in","r",stdin);
freopen("term.out","w",stdout);
n = read(), K = read();
mod = read();
if(n <= 1000) {
BF1::Main();
return 0;
}
return 0;
}
T3
思路
将所有的点点权全都放到边上去。然后建个反图，找出S可以到达的边集与T可以到达的边集的交集。然后将答案拆位从高到低贪心。用bitset优化一下。就可以在两秒里跑过了
代码
//2.424s
//2.211s
//2.121s
//2.013s
//1.917s
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<ctime>
#include<cstring>
#include<queue>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 200000 + 100,M = 500000 + 100;
bitset<M>s,t,tmp;
char buf[100000],*_p1 = buf,*_p2 = buf;
#define nc() (_p1==_p2&&(_p2=(_p1=buf)+fread(buf,1,100000,stdin),_p1==_p2) ? EOF :*_p1++)
inline int read() {
int x=0,f=1;
char ch=nc();
for(; !isdigit(ch); ch=nc())if(ch=='-')f=-1;
for (; isdigit(ch); ch=nc())x=x*10+ch-'0';
return x*f;
}
struct node {
int v,nxt;
}e[3][M];
int a[N];
int head[3][N],ejs[3];
int vis[N];
int n,m,S,T;
queue<int>q,q2;
inline void bfsS() {
while(!q.empty()) q.pop();
q.push(S);
memset(vis,0,sizeof(vis));
vis[S] = 1;
while(!q.empty()) {
int u = q.front();q.pop();
for(int i = head[1][u];i;i = e[1][i].nxt) {
int v = e[1][i].v;
s[i] = 1;
if(vis[v]) continue;
vis[v] = 1;
q.push(v);
}
}
}
inline void bfsT() {
while(!q.empty()) q.pop();
q.push(T);
memset(vis,0,sizeof(vis));
while(!q.empty()) {
int u = q.front();q.pop();
for(int i = head[2][u];i;i = e[2][i].nxt) {
int v = e[2][i].v;
t[i] = 1;
if(vis[v]) continue;
vis[v] = 1;
q.push(v);
}
}
}
int W[M];
int vis2[N],vis1[N];
inline void bfs1() {
while(!q.empty()) {
int u = q.front();q.pop();
for(int i = head[1][u];i;i = e[1][i].nxt) {
int v = e[1][i].v;
tmp[i] = 1;
if(vis1[v] == 1) continue;
vis1[v] = 1;
q.push(v);
}
}
}
inline void bfs2() {
while(!q2.empty()) {
int u = q2.front();q2.pop();
for(int i = head[2][u];i;i = e[2][i].nxt) {
int v = e[2][i].v;
tmp[i] = 1;
if(vis2[v] == 1) continue;
vis2[v] = 1;
q2.push(v);
}
}
}
int main() {
freopen("rabbit.in","r",stdin);
// freopen("rabbit.out","w",stdout);
n = read(),m = read(),S = read(),T = read();
for(int i = 1;i <= n;++i) a[i] = read();
for(int i = 1;i <= m;++i) {
int u = read(),v = read(),w = read();
e[1][i].v = v;e[1][i].nxt = head[1][u];head[1][u] = i;
e[2][i].v = u;e[2][i].nxt = head[2][v];head[2][v] = i;
W[i] = a[u] & a[v] & w;
}
if(S == T) {
printf("%d",a[S]);
return 0;
}
bfsS();
bfsT();
tmp = s & t;
if(!tmp.count()) {
puts("-1");return 0;
}
for(int i = 1;i <= m;++i) {
if(tmp[i] && W[i] == 0) {
puts("0");
return 0;
}
}
int ans = 0;
for(int i = 30;i >= 0;--i) {
memset(vis1,0,sizeof(vis1));
memset(vis2,0,sizeof(vis2));
int bz = 0;
for(int j = 1;j <= m;++j) {
if(tmp[j] == 0) continue;
if(!(W[j] & (1<<i))) {
bz = 1;
if(!vis1[e[1][j].v]) {
q.push(e[1][j].v);
vis1[e[1][j].v] = 1;
}
if(!vis2[e[2][j].v]) {
q2.push(e[2][j].v);
vis2[e[2][j].v] = 1;
}
}
}
if(bz == 0) {
ans |= (1<<i);
continue;
}
for(int j = 1;j <= m;++j)if((W[j] & (1 << i))) tmp[j] = 0;
bfs1();
bfs2();
tmp = tmp & s & t;
}
cout<<ans;
return 0;
}
总结
期望得分100 + 60 + 100 = 260
实际得分100 + 60 + 100 = 260
一言
越是代自己辩护，越是暴露自己的过错。 ——红与黑