SRM 453 DIV1 总结
250p: 直接暴力搜出所有情况。。。
500p:
题意：
有n支队伍进行积分制联赛，每两支队伍可以相互比赛任意次（包括0次），赢得队伍得w分，输的不得分，平局各得d分，现在给你一个积分榜，问最小需要多少场次能达到这个积分榜，如果不能达到，返回-1。
解题思路：
对于一个积分f，有w x + d y = f,根据扩展欧几里得，对于所有的积分都能求出最小的x，因为y = (y0*f/gcd - d/gcd *t)，对y排个序，y0, y1, y2, y3..., ，所有的y的和如果是奇数，方案肯定不可行，如果是偶数，要确保最大的数不大于其他所有数的和。
250p:
/* **********************************************
Author : JayYe
Created Time: 2013-8-13 9:15:42
File Name : final.cpp
*********************************************** */
#include <cstdio>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <sstream>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <bitset>
#include <algorithm>
using namespace std;
//typedef __int64 LL;
typedef long long LL;
class TheBasketballDivOne {
public:
int find(int n, int m) ;
};
int ans, vis[11], a[11];
bool mp[111111];
void dfs(int x, int y, int m, int n) {
int i;
if(x == n) {
for(i = 1;i <= n; i++) a[i] = vis[i];
sort(a+1, a+n+1);
if(a[n] == m) {
int sum = 0;
for(i = 1;i <= n; i++) sum = sum*10+a[i];
if(!mp[sum]) {
mp[sum] = 1;ans++;
}
}
return ;
}
if(y > n) {
dfs(x+1, x+2, m, n); return ;
}
vis[x] += 2;
dfs(x, y+1, m, n);
vis[x] --;
vis[y] ++;
dfs(x, y+1, m, n);
vis[x] --;
vis[y] ++;
dfs(x, y+1, m, n);
vis[y] -= 2;
}
int TheBasketballDivOne::find(int n, int m) {
ans = 0;
memset(vis, 0, sizeof(vis));
memset(mp, 0, sizeof(mp));
dfs(1, 2, m, n);
return ans;
}
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500p:
/* **********************************************
Author : JayYe
Created Time: 2013-8-13 15:22:44
File Name : final.cpp
*********************************************** */
#include <cstdio>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <sstream>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <bitset>
#include <algorithm>
using namespace std;
class TheTournamentDivOne {
public:
int find(vector <int> points, int w, int d) ;
};
int exgcd(int a, int b, int &x ,int &y) {
if(b == 0) {
x = 1; y = 0;
return a;
}
int ans = exgcd(b, a%b, y, x);
y -= a/b*x;
return ans;
}
int a[111111];
struct PP {
int val;
PP() {}
PP(int val) : val(val) {}
bool operator < (const PP &a) const {
return val < a.val;
}
}cur;
priority_queue<PP> q;
int TheTournamentDivOne::find(vector <int> ps, int w, int d) {
while(!q.empty()) q.pop();
int x, y, i;
int g = exgcd(w, d, x, y);
int extr = 0;
int sum = 0;
for(i = 0;i < ps.size(); i++) {
if(ps[i]%g != 0) return -1;
int xx = x*ps[i]/g;
xx = (xx%(d/g) + d/g)%(d/g);
int yy = (ps[i] - w*xx)/d;
if(yy < 0) return -1;
sum += yy;
extr += xx;
cur = PP(yy);
q.push( cur );
// printf("%d %d %d\n", ps[i], xx, yy);
}
int ret = -1;
// printf("sum = %d extr = %d\n", sum, extr);
while(!q.empty()) {
cur = q.top(); q.pop();
// printf("cur = %d\n", cur);
int val = cur.val;
if(sum %2 ==0 && val <= sum/2) {
if(ret == -1) ret = sum/2 + extr;
else if(ret > sum/2+extr)
ret = sum/2 + extr;
}
if(val >= w/g) {
cur.val = val - w/g;
extr += d/g;
sum -= w/g;
q.push(cur);
}
else break;
}
return ret;
}
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