题目很好，比较活跃，有难度，同时对基础要求又比较好，总体感觉这题目做起来挺不错的，没做出的去补一下
http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1318
数学题：
#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>
#define ll long long
#define eps 1e-8
#define inf 0xfffffff
//const ll INF = 1ll<<61;
using namespace std;
//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;
int main() {
ll n;
while(scanf("%lld",&n) == 1) {
ll ans = 1;
ll num = 1;
while((num*2) - 1 < n) {
ans++;
num *= 2;
}
printf("%lld\n",ans);
}
return EXIT_SUCCESS;
}
http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1321
spfa，记录最短路径的同时记录妹子的数量
#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>
#define ll long long
#define eps 1e-8
#define inf 0xfffffff
//const ll INF = 1ll<<61;
using namespace std;
//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;
int n,m;
bool vis[10000 + 5];
int dis[10000 + 5];
int ans[10000 + 5];
int num [10000 + 5];
int head[10000 + 5];
typedef struct Node {
int from,to;
int nex;
int w;
};
Node edge[100000 + 5];
int tot;
void clear() {
memset(num,0,sizeof(num));
memset(edge,0,sizeof(edge));
memset(head,-1,sizeof(head));
tot = 0;
}
void add(int u,int v,int w) {
edge[tot].from = u;
edge[tot].to = v;
edge[tot].nex = head[u];
edge[tot].w = w;
head[u] = tot++;
}
int spfa(int s,int e) {
memset(vis,false,sizeof(vis));
for(int i=0;i<=n;i++)
dis[i] = inf;
for(int i=1;i<=n;i++)
ans[i] = inf;
dis[s] = 0;
ans[s] = num[s];
vis[s] = true;
queue<int> q;
q.push(s);
while(!q.empty()) {
int u = q.front();
q.pop();
vis[u] = false;
for(int i=head[u];i!=-1;i=edge[i].nex) {
int v = edge[i].to;
if(dis[u] + edge[i].w < dis[v]) {
dis[v] = edge[i].w + dis[u];
ans[v] = ans[u] + num[v];
if(!vis[v]) {
vis[v] = true;
q.push(v);
}
}
else if(dis[v] == dis[u] + edge[i].w) {
if(ans[v] < ans[u] + num[v]) {
ans[v] = ans[u] + num[v];
if(!vis[v]) {
vis[v] = true;
q.push(v);
}
}
}
}
}
if(dis[n] == inf)
return -1;
else
return ans[n];
}
int main() {
while(scanf("%d %d",&n,&m) == 2) {
clear();
for(int i=1;i<=n;i++)
scanf("%d",&num[i]);
for(int i=1;i<=m;i++) {
int u,v,w;
scanf("%d %d %d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
int answer = spfa(1,n);
printf("%d\n",answer);
}
return EXIT_SUCCESS;
}http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1320
仔细写几个，不能遗漏，发现是卡特兰数，因为要取模 涉及到整除问题，所以要用扩展欧几里德求乘法逆元来避免这个问题
#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>
#define ll long long
#define eps 1e-8
#define inf 0xfffffff
//const ll INF = 1ll<<61;
using namespace std;
//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;
const ll MOD = 1000000007;
ll dp[10000 + 5];
ll n;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if(!b) {
x = 1; y = 0;
return a;
}
ll r = exgcd(b, a%b, y, x);
y -= a/b*x;
return r;
}
ll inv(ll a, ll m) //求逆元直接模版套上
{
ll x,y,gcd = exgcd(a, m, x, y);
if(x < 0)
x += m;
return x;
}
void clear() {
memset(dp,0,sizeof(dp));
dp[1] = 1;
dp[2] = 2;
for(int i=3;i<10001;i++) {
dp[i] = ((dp[i-1]*(4*i-2)%MOD) * ((inv(i+1,MOD) + MOD)%MOD))%MOD;
}
}
int main() {
clear();
while(scanf("%lld",&n) == 1) {
printf("%lld\n",dp[n]);
}
return EXIT_SUCCESS;
}符合分组背包，所以用分组背包做一下，注意分组要求
#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>
#define ll long long
#define eps 1e-8
#define inf 0xfffffff
//const ll INF = 1ll<<61;
using namespace std;
//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;
vector<int> G[1000 + 5];
int n,V,q;
int dp[1000 + 5];
int p[1000 + 5],w[1000 + 5];
int father[1000 + 5];
void clear() {
memset(dp,0,sizeof(dp));
for(int i=0;i<1005;i++)
father[i] = i;
for(int i=0;i<1005;i++)
G[i].clear();
}
int find(int x) {
if(father[x] != x)
return find(father[x]);
return x;
}
void merge(int x,int y) {
int dx = find(x);
int dy = find(y);
if(dx!=dy) {
father[dx] = dy;
}
}
int main() {
while(scanf("%d %d %d",&n,&V,&q) == 3) {
clear();
for(int i=0;i<n;i++)
scanf("%d %d",&p[i],&w[i]);
for(int i=0;i<q;i++) {
int tx,ty;
scanf("%d %d",&tx,&ty);
tx--;
ty--;
merge(tx,ty);
}
for (int i=0;i<n;i++) {
int x = find(i);
G[x].push_back(i);
}
for (int i=0;i<n;i++) {
if (G[i].size() != 1)
continue;
for (int j=V;j>=w[i];j--)
if (dp[j - w[i]] + p[i] > dp[j])
dp[j] = dp[j - w[i]] + p[i];
}
for (int i=0;i<n;i++) {
if (G[i].size() == 1)
continue;
for (int k=V;k>=0;k--) {
for (int j=0;j<G[i].size();j++) {
if (w[G[i][j]] > k)
continue;
if (dp[k - w[G[i][j]]] + p[G[i][j]] > dp[k])
dp[k] = dp[k - w[G[i][j]]] + p[G[i][j]];
}
}
}
printf("%d\n",dp[V]);
}
return EXIT_SUCCESS;
}
CSU Monthly 2013 Oct 中南大学ACM月赛,布布扣,bubuko.com
CSU Monthly 2013 Oct 中南大学ACM月赛
原文：http://blog.csdn.net/yitiaodacaidog/article/details/20860695