标签：acm   hdu   贪心   bestcoder   

Go to movies

3 2
2 2
3 5

4

Hint
LeLe can buy four tickets with four yuan in cinema 1.
 

思路：简单贪心

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream> 
#include <algorithm>
#include <cmath>
#define INF 0x3fffffff
using namespace std;

struct node {
	int a, b;
}dian[105];

int main() {
	int n, m;
	while(scanf("%d %d", &n, &m) != EOF) {
		for(int i = 0; i < m; i++) {
			scanf("%d %d", &dian[i].a, &dian[i].b);
		}
		
		int ans = INF;
		for(int i = 0; i<m; i++) {
			int t = n / dian[i].a;
			if(n % dian[i].a) t++;
			if(t * dian[i].b < ans) ans = t * dian[i].b;
		}
		printf("%d\n", ans);
	}
	return 0;
} 

Building Blocks

4 3 2
1 2 3 5
4 4 4
1 2 3 4

1
-1

Hint
In first case, LeLe move one block from third pile to first pile.
 

这一场真是太爽了，简直无法忍受了，一直错，，，明明想法是好的，每次都有小BUG，我他妈真是醉了，写得真挫，

思路：其实很简单，就是算出长度为W的区间的正数之和n和负数之和，负数之和再取个绝对值得m，取max(n,m)为MIN，再往后扫每一段长度为W的区间，取其中最小值

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#define INF 0x7fffffff
using namespace std;

typedef long long LL;
const int maxn = 50005;
LL mu[maxn * 3];
LL zhen;
LL fu; 

int main() {
	int n, W, H;
	while(scanf("%d%d%d", &n, &W, &H) != EOF) {
		LL sum = 0;
		for(int i = W; i < W + n; i++) {
			scanf("%I64d", &mu[i]);
			sum += mu[i];
		}
		if(sum < LL(H) * W) {
			printf("-1\n");
		}
		else 
		{
			LL MIN, m = n + W * 2;
			for(int i = 0; i < W; i++) mu[i] = 0;
			for(int i = W + n; i < m; i++) mu[i] = 0;
			zhen = 0; fu = LL(H) * W;
			MIN = max(zhen, fu);
			for(int i = W; i < m; i++) {
				if(mu[i - W] < H) fu -= (H - mu[i - W]);
				else zhen -= mu[i - W] - H;
				if(mu[i] < H) fu += (H - mu[i]);
				else zhen += mu[i] - H;
				MIN = min(MIN, max(zhen, fu));
			}
			printf("%I64d\n", MIN);
		}
	}
	return 0;
}

一直让我魂牵梦绕调BUG的代码（尼玛！）：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#define INF 0x7fffffff
using namespace std;

typedef long long LL;
const int maxn = 50005;
LL mu[maxn * 3];
LL zhen;
LL fu; 

int main() {
    int n, W, H;
    while(scanf("%d%d%d", &n, &W, &H) != EOF) {
        LL sum = 0;
        memset(mu, 0, sizeof(mu));
        for(int i = W; i < W + n; i++) {
            scanf("%I64d", &mu[i]);
            mu[i] -= H;
            sum += mu[i];//其实这里可以放上面，我就不用找的这么辛苦了。。。哎。。。代码能力。 
        }
        if(sum < 0) {			//错误！这里sum值不是判断其小于0，而是判断sum+n*H<W*H,
            printf("-1\n");		//我靠，劳资就是这里被自己误导的，
        }						//写程序真的是要严谨严谨再严谨啊！！！！
        else 
        {
            LL MIN, m = n + W * 2;
            for(int i = 0; i < W; i++) mu[i] -= H;
            for(int i = W + n; i < m; i++) mu[i] -= H;
            zhen = 0; fu = LL(H) * W;
            MIN = max(zhen, fu);
            for(int i = W; i < m; i++) {
                if(mu[i - W] < 0) fu -= (-mu[i - W]);
                else zhen -= mu[i - W];
                if(mu[i] < 0) fu += (-mu[i]);
                else zhen += mu[i];
                MIN = min(MIN, max(zhen, fu));
            }
            printf("%I64d\n", MIN);
        }
    }
    return 0;
}

HDU - 5190 - Go to movies && 5191 - Building Blocks （BC#34 A,B）

标签：acm   hdu   贪心   bestcoder   

原文地址：http://blog.csdn.net/u014355480/article/details/44524337