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  • Codeforces 313

    A. Ilya and Bank Account 水题

    B. Ilya and Queries 水,维护区间和

    C. Ilya and Matrix

    贪心。大的明显会被多次取。

    /**
     * Problem:313C
     * Author:Shun Yao
     * Time:2013.5.31
     * Result:Accepted
     * Memo:greedy
     */
    
    #include <cstdio>
    #include <algorithm>
    #include <functional>
    
    const long MaxN = 2000001;
    
    long n, N, a[MaxN];
    long long s[MaxN], ans;
    
    int main() {
    	static long i;
    	scanf("%ld", &N);
    	for (i = 1; i <= N; ++i)
    		scanf("%ld", a + i);
    	std::sort(a + 1, a + N + 1, std::greater<long>());
    	s[0] = 0;
    	for (i = 1; i <= N; ++i)
    		s[i] = s[i - 1] + a[i];
    	ans = 0;
    	for (i = 1; i <= N; i <<= 2)
    		ans += s[i];
    	printf("%lld", ans);
    	return 0;
    }
    

     D. Ilya and Roads

    好吧,dp。。。

     状态很好想,f[i][j]代表前i个人覆盖了j个。

    然后是g[i][j]代表覆盖区间 i -- j 的最小cost;

    转移见code:

    /**
     * Problem:313D
     * Author:Shun Yao
     * Time:2013.6.3
     * Result:Accepted
     * Memo:DP
     */
    
    #include <cstdio>
    
    long long min(long long x, long long y) {
    	return x < y ? x : y;
    }
    
    const long Maxn = 305;
    const long long inf = 1LL << 60;
    
    long n, m, K;
    long long g[Maxn][Maxn], f[Maxn][Maxn], ans;
    
    int main() {
    	static long a, b, c, i, j, k;
    	scanf("%ld%ld%ld", &n, &m, &K);
    	for (i = 0; i <= n; ++i)
    		for (j = 0; j <= n; ++j)
    			g[i][j] = inf;
    	for (i = 1; i <= m; ++i) {
    		scanf("%ld%ld%ld", &a, &b, &c);
    		g[a][b] = min(g[a][b], c);
    	}
    	for (i = 1; i <= n; ++i)
    		for (j = i; j <= n; ++j)
    			g[i][j] = min(g[i][j], g[i - 1][j]);
    	for (i = 0; i <= n; ++i) {
    		f[i][0] = 0;
    		for (j = 1; j <= n; ++j)
    			f[i][j] = inf;
    	}
    	for (i = 1; i <= n; ++i)
    		for (j = 0; j <= i; ++j) {
    			f[i][j] = f[i - 1][j];
    			for (k = i - j + 1; k <= i; ++k)
    				if (f[k - 1][j - (i - k + 1)] != inf && g[k][i] != inf)
    					f[i][j] = min(f[i][j], f[k - 1][j - (i - k + 1)] + g[k][i]);
    		}
    	ans = inf;
    	for (i = K; i <= n; ++i)
    		if (ans > f[n][i])
    			ans = f[n][i];
    	printf("%I64d", ans == inf ? -1 : ans);
    	return 0;
    }
    

    E. Ilya and Two Numbers

    贪心,构造。

    /**
     * Problem:313E
     * Author:Shun Yao
     * Time:2013.6.15
     * Result:Acceped
     * Memo:constructive algorithms, greedy
     */
    
    #include <cstring>
    #include <cstdio>
    #include <functional>
    #include <algorithm>
    
    #define Maxn 100005
    
    long n, m, a[Maxn], b[Maxn], c[Maxn], d[Maxn], ans[Maxn];
    
    int main() {
    	static long i, x, l;
    #ifndef ONLINE_JUDGE
    	freopen("e.in", "r", stdin);
    	freopen("e.out", "w", stdout);
    #endif
    	memset(a, 0, sizeof a);
    	memset(b, 0, sizeof b);
    	scanf("%ld%ld", &n, &m);
    	for (i = 1; i <= n; ++i) {
    		scanf("%ld", &x);
    		++a[x];
    	}
    	for (i = 1; i <= n; ++i) {
    		scanf("%ld", &x);
    		++b[x];
    	}
    	ans[0] = 0;
    	d[0] = 0;
    	for (i = m - 1; i >= 0; --i) {
    		while (b[m - 1 - i]--)
    			c[++l] = m - 1 - i;
    		while (a[i]--) {
    			if (l)
    				ans[++ans[0]] = i + c[l--];
    			else
    				d[++d[0]] = i;
    		}
    	}
    	for (i = 1; i <= d[0]; ++i)
    		ans[++ans[0]] = d[i] + c[l--] - m;
    	std::sort(ans + 1, ans + n + 1, std::greater<long>());
    	for (i = 1; i <= n; ++i)
    		printf("%ld ", ans[i]);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/hsuppr/p/3118614.html
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