The 2012 Rocky Mountain Regional Contest 题解

A:

UVALive 6084

Happy Camper

#include <cstdio>
typedef long long ll;
int main(){
	ll l, v, p;
	int cas = 0;
	while(~scanf("%lld%lld%lld", &l, &p, &v)) {
		if(l == 0 && p == 0 && v == 0) break;
		ll ans;
		if(v%p > l) ans = v/p*l+l;
		else ans = v/p*l+ v%p;
		
		printf("Case %d: %lld\n", ++cas, ans);
	}
    return 0;
}

B:

UVALive 6085

Chemistry

题意：

按字典序把每个元素的个数都输出来。

维护个栈，跟括号匹配差不多，写写写。。

#include <cstdio>
#include <iostream>
#include <string>
#include <map>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
typedef map<string, ll> M;
typedef pair<string, ll> pii;
const int N =100+10;
char s[N];
vector<pii> out;

void merge(M& a, M b, ll v) {
	for (M::iterator it = b.begin(); it != b.end(); ++it)
		a[it->first] += v*it->second;
}
M cal(char *s, int len) {
	M re;
	for (int i = 0; i < len;) {
		if (s[i] == '(') {
			int k = 0, j;
			ll v = 1;
			for (j = i; j < len; ++j) {
				if (s[j] == '(')
					++ k;
				if (s[j] == ')') {
					if (--k == 0)
						break;
				}
			}
			M g = cal(s+i+1, j-i-1);
			++ j;
			if (s[j] >= '0' && s[j] <= '9') {
				v = 0;
				for (; j < len; ++j)
					if (s[j] >= '0' && s[j] <= '9') {
						v = v * 10 + s[j] - '0';
					} else break;
			}
			merge(re, g, v);
			i = j;
		} else {
			string ch = "";
			ch += s[i];
			int j;
			ll v = 1;
			for (j = i + 1; j < len; ++j) {
				if (s[j] >= 'a' && s[j] <= 'z')
					ch += s[j];
				else break;
			}
			if (s[j] >= '0' && s[j] <= '9') {
				v = 0;
				for (; j < len; ++j) {
					if (s[j] >= '0' && s[j] <= '9')
						v = v * 10+ s[j]-'0';
					else break;
				}
			}
			re[ch] += v;
			i = j;
		}
	}
	return re;
}
void work() {
	int len = strlen(s);
	M ans = cal(s, len);
	out.clear();
	for (M::iterator it = ans.begin(); it != ans.end(); ++it)
		out.push_back(pii(it->first, it->second));
	sort(out.begin(), out.end());
	int f = 0;
	for (int i = 0; i < out.size(); ++i) {
		if (f)
			putchar('+');
		if (out[i].second > 1)
			printf("%lld", out[i].second);
		if (out[i].second > 0) {
			f = 1;
			cout << out[i].first ;
		}
	}
	puts("");
}
int main() {
	while (~scanf("%s", s))
		work();
	return 0;
}

D:

UVALive 6087

Fuel Stops

题意：

给定n 个加油站

下面n 个数表示每个 加油站能给车子加的油量

下面n个数表示这个加油站开往下一个加油站需要的油量 i->i+1, n->1

问 ：

从哪些加油站出发，是不需要额外用油的（即只需要使用加油站提供的油量）

油箱容量无限

思路：

见代码

#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N =100000+5;
int T = 0, n, a[N], b[N], ans[N], dep;
ll vr[N], mir[N], mil[N];

void work() {
	for (int i = 1; i <= n; ++i)
		scanf("%d", &a[i]);
	for (int i = 1; i <= n; ++i)
		scanf("%d", &b[i]);
	ll v = 0, mi = 0;
	for (int i = 1; i <= n; ++i) {
		if (v<mi)
			mi = v;
		mil[i] = mi;//计算从1->i过程中出现的最小油量(可能<0)
		v += a[i];
		v -= b[i];
	}
	vr[n+1] = mir[n+1] = 0;//n+1就等价于1这个点
	for (int i = n; i >= 1; --i) {
		v = a[i];
		v -= b[i];
		vr[i] = vr[i+1] + v;//i 到达 1后剩余油量
		mir[i] = min(mir[i+1]+v, v);//计算从i->1过程中出现的最小油量  i->1的最小油量= (i+1)->1的最小油量 + v
	}
	dep = 0;
	for (int i = 1; i <= n; ++i)
		if (mir[i] >= 0)
			if (mil[i] + vr[i] >= 0) {//有了剩余油量就相当于全程的最小值都+vr[i] ，这样就能得到全程最小值
				ans[dep++] = i;
			}

	printf("Case %d:", ++T);
	for (int i = 0; i < dep; ++i)
		printf(" %d", ans[i]);
	puts("");
}
int main() {
	while (~scanf("%d", &n), n) 
		work();
	return 0;
}

E:

UVALive 6088

Approximate Sorting

题解：点击打开链接

F: 

UVALive 6089

Nine

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <bits/stdc++.h>
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != '-' && (c<'0' || c>'9')) c = getchar();
	sgn = (c == '-') ? -1 : 1;
	ret = (c == '-') ? 0 : (c - '0');
	while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x <0) {
		putchar('-');
		x = -x;
	}
	if (x>9) pt(x / 10);
	putchar(x % 10 + '0');
}
using namespace std;
typedef long long ll;
const int N = 505;
struct node{
    int jiu, cha, x, y;
    node(int a=0,int b=0,int c=0,int d=0):jiu(a),cha(b),x(c),y(d){}
}ans;
node work(node X, node Y){
    bool hehe;
    if(X.jiu != Y.jiu){
        hehe = X.jiu > Y.jiu;
    }
    else if(X.cha != Y.cha){
        hehe = X.cha < Y.cha;
    }
    else if(X.x != Y.x){
        hehe = X.x < Y.x;
    }
    else hehe = X.y < Y.y;
    return hehe ? X : Y;
}
int Find(int a, int b){
    int siz = 0;
    while(a){
        siz += (a%10)==9;
        a/=10;
    }
    while(b){
        siz += (b%10)==9;
        b/=10;
    }
    return siz;
}
int h, s, all;
int C(int a, int b){
    int siz = a*60+b;
    siz = all - siz;
    if(siz < 0) siz = -siz;
    return siz;
}
int main(){
    while (~scanf("%d:%d", &h,&s), h+s){
        all = h*60+s;
        ans =  node(Find(h,s), 0, h, s);
        for(int i = 0; i < 100; i++)
            for(int j = 0; j < 100; j++)
            {
                node tmp = node(Find(i,j), C(i,j), i, j);
                if(tmp.cha * 10 < all)
                {
                    ans = work(ans, tmp);
                }
            }
        printf("%02d:%02d\n", ans.x, ans.y);
	}
	return 0;
}

H:

UVALive 6091

Trees

题解：点击打开链接

I:

UVALive 6092

Catching Shade in Flatland

题解:点击打开链接