17国庆day3

2015-2016 Northwestern European Regional Contest (NWERC 2015)

Elementary Math

 Gym - 101485E 

题意:给n组数, 可以加三种(加减乘)运算, 使得所有的结果都不相同. 如果不存在则输出impossible.

二分图匹配, 每一组数像不同的结果连边, 求最大匹配,如果最大匹配等于n则有解.

其实还是比较裸的,场上只想到暴力=_=

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxe = 8000;
const int maxn = 2510;

map<ll,int> mp;
ll A[maxe];
ll p1[maxn], p2[maxn];

struct Edge{
    int u,v,nex;
    Edge(int u=0,int v=0,int nex=0): u(u), v(v), nex(nex) {}
}e[maxe];
int head[maxn];
int cnt;
void init(){
    memset(head, -1, sizeof(head));
    cnt = 0;
}
void add(int u,int v) {
    e[cnt] = Edge(u,v,head[u]);
    head[u] = cnt++;
}

int vg[maxn], vb[maxe], mcg[maxn], mcb[maxe];

int Hungary(int u){
    vg[u] = 1;
    for(int i = head[u]; ~i; i = e[i].nex) {
        int v = e[i].v;
        if(!vb[v]) {
            vb[v] = 1;
            if(mcb[v] == -1 || Hungary(mcb[v])){
                mcb[v] = u;
                mcg[u] = v;
                return 1;
            }
        }
    }
    return 0;
}

int main() {
    int n;
    while(scanf("%d", &n)!=EOF) {
        init();
        mp.clear();
        ll a, b;
        int id = 0;
        for(int i = 0; i < n; i++) {
            scanf("%lld %lld", &a, &b);
            p1[i] = a;
            p2[i] = b;
            ll u = a+b;
            if(!mp.count(u)) {
                A[id] = u;
                mp[u] = id++;
            }
            add(i, mp[u]);
            u = a-b;
            if(!mp.count(u)) {
                A[id] = u;
                mp[u] = id++;
            }
            add(i, mp[u]);
            u = a*b;
            if(!mp.count(u)) {
                A[id] = u;
                mp[u] = id++;
            }
            add(i, mp[u]);
        }
        int ans = 0;
        memset(mcb, -1, sizeof(mcb));
        memset(mcg, -1, sizeof(mcg));
        for(int i = 0; i < n; i++) {
            memset(vb, 0, sizeof(vb));
            memset(vg, 0, sizeof(vg));
            if(Hungary(i)) ans++;
        }
       // cout<<ans<<endl;
        if(ans != n) {
            puts("impossible");
            continue;
        }
        for(int i = 0; i < n; i++) {
            int v = mcg[i];
            if(p1[i]+p2[i] == A[v]) {
                printf("%lld + %lld = %lld
", p1[i], p2[i], A[v]);
                continue;
            } else if(p1[i]-p2[i] == A[v]){
                printf("%lld - %lld = %lld
", p1[i], p2[i], A[v]);
                continue;
            } else {
                printf("%lld * %lld = %lld
", p1[i], p2[i], A[v]);
                continue;
            }
        }
    }
    return 0;
}

Debugging

 Gym - 101485D 

题意:n行代码,有一个bug,可以输出中间结果debug, 每加一行printf语句耗时b, 编译一次耗时a,问最短多长时间找到bug.

dp[n]表示n行需要的时间, 暴力把n行分成2到n段,取最小的时间就是结果.

刚看到这道题的时候以为和高楼抛水球那个题一样......没绕出来......

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 1e6+10;
const ll inf = 1e16;
ll d[maxn];
ll n, a, b;

ll solve(ll n) {
    if(n <= 1) return 0;
    if(d[n]) return d[n];
    ll res = inf;
    for(ll i = 2; i <= n; i++) {
        res = min(res, solve(ceil((double)n/i))+b*(i-1)+a);
    }
    return d[n]=res;
}

int main() {
    while(scanf("%lld %lld %lld", &n, &a, &b)!=EOF) {
        printf("%lld
", solve(n));
    }
}