2020CSP-J2比赛记录&题解

题目请看洛谷

备注：这次比赛我是没打的

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T1

先把数转成二进制，逐位计算，并判断是否可完整正确拆分

贴一下代码

#include <bits/stdc++.h>
using namespace std;
#define fre(c) freopen(c".in","r",stdin);freopen(c".out","w",stdout);
#define ll long long
#define endl "\n"
#define ios ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define cst const
cst ll N = 1e3 + 5;
ll n, a[N], cnt, now = 1;
ll read() {
	char c;
	ll sum = 0, f = 1;
	c = getchar();
	while (!isdigit(c)) {
		if (c == '-')
			f *= -1;
		c = getchar();
	}
	while (isdigit(c)) {
		sum = sum * 10 + (c - '0');
		c = getchar();
	}
	return sum * f;
}
int main() {
//	ios
//	fre("")
	n = read();
	if (n % 2 == 1) {cout << -1;return 0;}
	now = 1;
	while (now <= n) now *= 2;
	while (n > 0) {	if (now <= n) {	a[cnt ++] = now; n -= now; } now /= 2; }
	for (int i = 0; i < cnt; i ++) { if (i > 0) {cout << " ";} cout << a[i]; }
	return 0;
}

赛时：-point

赛后：100point

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T2

以前看过思路

用个桶存，无脑sort也可以拿分，存储分数线低的位置，听说还可以用对顶堆

贴一下代码

#include <bits/stdc++.h>
using namespace std;
#define fre(c) freopen(c".in","r",stdin);freopen(c".out","w",stdout);
#define ll long long
#define endl "\n"
#define ios ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define cst const
cst ll N = 1e3 + 5;
ll n, w, x, a[N], sum;
ll read() {
	char c;
	ll sum = 0, f = 1;
	c = getchar();
	while (!isdigit(c)) {
		if (c == '-')
			f *= -1;
		c = getchar();
	}
	while (isdigit(c)) {
		sum = sum * 10 + (c - '0');
		c = getchar();
	}
	return sum * f;
}
int main() {
//	ios
//	fre("")
	n = read();
	w = read();
	for (ll i = 1; i <= n; i ++) {
		x = read();
		a[x] ++;
		sum = 0;
		for (ll j = 600; j >= 0; j --) {
			sum += a[j];
			if (sum >= max(1LL, i * w / 100)) {
				cout << j << " ";
				break ;
			}
		}
	}
	return 0;
}

赛时：-point

赛后：100point

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T3

先用转成栈，接着变成树形结构

贴一下标程

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const LL mod = 1e9 + 7;
const int N = 1000005;

char s[N];
int a[N];
int son[N][2], ck;
int flag[N], c[N];
int n, q;
int dfs(int u, int g) {
    a[u] ^= g;
    if (u <= n) {
        return a[u];
    }
    int x = dfs(son[u][0], g ^ flag[son[u][0]]);
    int y = dfs(son[u][1], g ^ flag[son[u][1]]);
    if (a[u] == 2) {
        if (x == 0) c[son[u][1]] = 1;
        if (y == 0) c[son[u][0]] = 1;
        return x & y;
    } else {
        if (x == 1) c[son[u][1]] = 1;
        if (y == 1) c[son[u][0]] = 1;
        return x | y;
    }
}
void dfs2(int u) {
    if (u <= n) return;
    c[son[u][0]] |= c[u];
    c[son[u][1]] |= c[u];
    dfs2(son[u][0]);
    dfs2(son[u][1]);
}
int main() {
    // freopen("expr.in", "r", stdin);
    // freopen("expr.out", "w", stdout);
    gets(s);
    scanf("%d", &n);
    ck = n;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
    stack<int> b;
    for (int i = 0; s[i]; i += 2) {
        if (s[i] == 'x') {
            int x = 0;
            i++;
            while (s[i] != ' ') {
                x = x * 10 + s[i] - '0';
                i++;
            }
            i--;
            b.push(x);
        } else if (s[i] == '&') {
            int x = b.top();
            b.pop();
            int y = b.top();
            b.pop();
            b.push(++ck);
            a[ck] = 2;
            son[ck][0] = x;
            son[ck][1] = y;
        } else if (s[i] == '|') {
            int x = b.top();
            b.pop();
            int y = b.top();
            b.pop();
            b.push(++ck);
            a[ck] = 3;
            son[ck][0] = x;
            son[ck][1] = y;
        } else if(s[i] == '!'){
            flag[b.top()] ^= 1;
        }
    }
    int ans = dfs(ck, flag[ck]);
    dfs2(ck);
    scanf("%d", &q);
    while (q--) {
        int x;
        scanf("%d", &x);
        printf("%d\n", c[x] ? ans : !ans);
    }
    return 0;
}

赛时：-point

赛后：-point

T4

记忆化搜索或dp，分析来时路

贴一下代码

#include <bits/stdc++.h>
using namespace std;
#define fre(c) freopen(c".in","r",stdin);freopen(c".out","w",stdout);
#define ll long long
#define endl "\n"
#define ios ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define cst const
cst ll N = 1e3 + 5;
ll n, m, a[N][N], dp[N][N];
inline ll read() {
	char c;
	ll sum = 0, f = 1;
	c = getchar();
	while (!isdigit(c)) {
		if (c == '-')
			f *= -1;
		c = getchar();
	}
	while (isdigit(c)) {
		sum = sum * 10 + (c - '0');
		c = getchar();
	}
	return sum * f;
}
int main() {
//	ios
//	#ifdef debug
//		fre("D")
//	#endif
	n = read();
	m = read();
	for (ll i = 1; i <= n; i ++) {
		for (ll j = 1; j <= m; j ++) {
			a[i][j] = read();
		}
	}
	for (ll i = 1; i <= n; i ++) {
		dp[i][1] = dp[i - 1][1] + a[i][1];
	}
	for (ll j = 2; j <= m; j ++) {
		vector<ll> left(n + 2), right(n + 2);
		left[1] = dp[1][j - 1] + a[1][j];
		for (ll i = 2; i <= n; i ++) {
			left[i] = max(left[i - 1], dp[i][j - 1]) + a[i][j];
		}
		right[n] = dp[n][j - 1] + a[n][j];
		for (ll i = n - 1; i >= 1; i --) {
			right[i] = max(right[i + 1], dp[i][j - 1]) + a[i][j];
		}
		for (ll i = 1; i <= n; i ++) {
			dp[i][j] = max(left[i], right[i]);
		}
	}
	cout << dp[n][m];
	return 0;
}

赛时：-point

赛后：100point

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