Codeforces 920D Tanks (看题解)

Tanks

最关键的一点就是怎么判方案是否存在。。 只要存在若干个坦克之和的sum % k == v % k 就有解， 否则无解。

我怎么想不到呢。。。 

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 5000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}


int n, k, v, sum, now1, now2, a[N];
bool dp[N][N];
bool f[N][N];
vector<int> vc[2];

int main() {
    scanf("%d%d%d", &n, &k, &v);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        sum += a[i];
    }
    if(sum < v) return puts("NO"), 0;
    dp[0][0] = true;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < k; j++) {
            if(!dp[i][j]) continue;
            dp[i + 1][(j + a[i + 1]) % k] = true;
            f[i + 1][(j + a[i + 1]) % k] = true;
            dp[i + 1][j] = true;
            f[i + 1][j] = false;
        }
    }
    if(!dp[n][v % k]) return puts("NO"), 0;
    int tmp = v % k;
    for(int i = n; i >= 1; i--) {
        if(f[i][tmp]) {
            vc[0].push_back(i);
            tmp = ((tmp - a[i]) % k + k) % k;
        } else {
            vc[1].push_back(i);
        }
    }
    sort(ALL(vc[0]));
    sort(ALL(vc[1]));
    for(auto& t : vc[0]) now1 += a[t];
    now2 = sum - now1;
    puts("YES");
    for(int i = 1; i < SZ(vc[0]); i++) printf("100000 %d %d
", vc[0][i], vc[0][0]);
    for(int i = 1; i < SZ(vc[1]); i++) printf("100000 %d %d
", vc[1][i], vc[1][0]);
    if(!SZ(vc[1])) vc[1].push_back(vc[0].back());
    if(!SZ(vc[0])) vc[0].push_back(vc[1].back());
    if(now1 < v) printf("%d %d %d
", (v - now1) / k, vc[1][0], vc[0][0]);
    if(now1 > v) printf("%d %d %d
", (now1 - v) / k, vc[0][0], vc[1][0]);
    return 0;
}

/*
*/