Codeforces 1244C

2021-03-13 12:05:28 阅读：182 来源： 互联网

标签：ch EOF int void Codeforces hasRead 1244C include

题意描述

\(x+y+z=n\)
\(xw+yd=p\)
求是否有满足方程的\((x,y,z)\)，没有输出\(-1\)

思路1（枚举）

由于\(d<w\)，平局的分数是严格小于获胜的分数。当\(x=d\)且\(y=w\)时，获胜的分数和平局的分数相同。假设平局的次数为\(y\)，当\(y<w\)时，枚举\([0,w-1]\)区间，寻找是否有答案。当\(y≥w\)时，平局得到的分数就变成\(d(w+i)=dw+di\)，获胜的场数就变成w+d,平局的场数就又变成了\([0,w-1]\)区间。所以我们可以枚举平局的次数，判断是否满足条件即可

思路2（exgcd）

待补

AC代码1

#include "iostream"
#include "cstring"
#include "string"
#include "vector"
#include "cmath"
#include "algorithm"
#include "map"
#include "set"
#include "queue"
#include "stack"
#include "cassert"
#include "unordered_map"
#include "sstream"
#include "cstdio"

using namespace std;

#define fi first
#define se second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) a.begin(),a.end()
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);
#define seteps(N) setprecision(N)
#define uni(x) sort(all(x)), x.erase(unique(all(x)), x.end())
#define lson (ind<<1)
#define rson (ind<<1|1) 
#define endl '\n'
#define dbg(x) cerr << #x " = " << (x) << endl
#define mp make_pair
//#define LOCAL

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 lll;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
typedef pair<int,PII> PIII;


struct Scanner {
 
    bool hasNext = 1;
    bool hasRead = 1;
 
    int nextInt() {
        hasRead = 0;
        int res = 0;
        char flag = 1, ch = getchar();
        while(ch != EOF && !isdigit(ch)) {
            hasRead = 1;
            flag = (ch == '-') ? -flag : flag;
            ch = getchar();
        }
        while(ch != EOF && isdigit(ch)) {
            hasRead = 1;
            res = res * 10 + (ch - '0');
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return res * flag;
    }
 
    ll nextLL() {
        hasRead = 0;
        ll res = 0;
        char flag = 1, ch = getchar();
        while(ch != EOF && !isdigit(ch)) {
            hasRead = 1;
            flag = (ch == '-') ? -flag : flag;
            ch = getchar();
        }
        while(ch != EOF && isdigit(ch)) {
            hasRead = 1;
            res = res * 10 + (ch - '0');
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return res * flag;
    }
 
    char nextChar() {
        hasRead = 0;
        char ch = getchar();
        while(ch != EOF && isspace(ch)) {
            hasRead = 1;
            ch = getchar();
        }
        if(ch == EOF)
            hasNext = 0;
        return ch;
    }
 
    int nextString(char *str) {
        hasRead = 0;
        int len = 0;
        char ch = getchar();
        while(ch != EOF && isspace(ch)) {
            hasRead = 1;
            ch = getchar();
        }
        while(ch != EOF && !isspace(ch)) {
            hasRead = 1;
            str[++len] = ch;
            ch = getchar();
        }
        str[len + 1] = 0;
        if(ch == EOF)
            hasNext = 0;
        return len;
    }
 
} sc;
 
ll rd() {
    ll x = sc.nextLL();
    return x;
}
 
void rd(int &x) {
    x = sc.nextInt();
}
 
void rd(ll &x) {
    x = sc.nextLL();
}
 
void rd(char &x) {
    x = sc.nextChar();
}
 
void rd(char* x) {
    sc.nextString(x);
}
 
template<typename T1, typename T2>
void rd(pair<T1, T2> &x) {
    rd(x.first);
    rd(x.second);
}
 
template<typename T>
void rd(T *x, int n) {
    for(int i = 1; i <= n; ++i)
        rd(x[i]);
}

template<typename T>
void rd(vector<T> &x,int n){
    for(int i = 1; i <= n; ++i)
        rd(x[i]);
}
 
void printInt(int x) {
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x >= 10)
        printInt(x / 10);
    putchar('0' + x % 10);
}
 
void printLL(ll x) {
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x >= 10)
        printLL(x / 10);
    putchar('0' + x % 10);
}
 
void pr(int x, char ch = '\n') {
    printInt(x);
    putchar(ch);
}
 
void pr(ll x, char ch = '\n') {
    printLL(x);
    putchar(ch);
}

template<typename T1, typename T2>
void pr(pair<T1, T2> x, char ch = '\n') {
#ifdef LOCAL
    putchar('<');   
    pr(x.first, ' ');
    pr(x.second, '>');
    putchar(ch);
    return;
#endif //LOCAL
    pr(x.first, ' ');
    pr(x.second, ch);
}
template<typename T>
void pr(T *x, int n) {
    for(int i = 1; i <= n; ++i)
        pr(x[i], " \n"[i == n]);
}
 
template<typename T>
void pr(vector<T> &x) {
    int n = x.size();
    for(int i = 1; i <= n - 1; ++i)
        pr(x[i], " \n"[i == n - 1]);
}

const int N=105;
const int M=1<<12;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll oone=1;
const double eps=1e-6;
const double pi=acos(-1);

ll n,p,d,w;
struct Solver
{
    void InitOnce(){

    }

    void Read(){
        rd(n);
        rd(p);
        rd(w);
        rd(d);
    }

    void Solve(){
        rep(i,0,w){
            ll cur=p-i*d;
            if(cur%w || cur<0) continue;
            ll win=cur/w;
            if(win+i<=n){
                pr(win,' ');
                pr(i,' ');
                pr(n-win-i);
                return;
            }
        }
        pr(-1);
    }
}solver;
int main(){
#ifdef LOCAL
    freopen("data.in","r",stdin);
#endif  //LOCAL
    solver.InitOnce();
    int t=1;
    //t=sc.nextInt();
    //t=INF;
    while(t--){
        solver.Read();
        if(!sc.hasRead) break;
        solver.Solve();
        if(!sc.hasNext) break;
    }
}

AC代码2

待补

标签：ch,EOF,int,void,Codeforces,hasRead,1244C,include
来源： https://www.cnblogs.com/dejavuzz/p/14528202.html