USACO 之 Section 1.1 Ad Hoc Problems

Your Ride Is Here：模擬一下題意過程，就可以了。。。

/*
ID: Jming
PROG: ride
LANG: C++
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
using namespace std;

int Solve(string str)
{
    int sum = 1;
    for (int i = 0; i < str.size(); i++) {
        sum *= (str[i] - 'A' + 1);
        sum %= 47;
    }
    return sum;
}

int main()
{
    freopen("ride.in", "r", stdin);
    freopen("ride.out", "w", stdout);
    string str;
    string str1;
    cin >> str >> str1;
    int tmp1, tmp2;
    tmp1 = Solve(str);
    tmp2 = Solve(str1);
    if (tmp1 == tmp2) {
        printf("GO\n");
    }else {
        printf("STAY\n");
    }
    return 0;
}

Greedy Gift Givers：用map即可解決。。。

/*
ID: Jming
PROG: gift1
LANG: C++
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <map>
using namespace std;
const int MAX_N = 15;
string arr_str[MAX_N];

int N;

struct node {
    int rec, giv;
    node() {
        rec = 0;
        giv = 0;
    }
};
node myrecord[MAX_N];
map<string, node> str_no_map;

void Solve() {
    string str, t_str;
    int amount, number;
    while (cin >> str) {
        scanf("%d %d", &amount, &number);
        if (0 == number) {
            continue;
        }
        str_no_map[str].giv += (amount / number) * number;
        for (int i = 0; i < number; ++i) {
            cin >> t_str;
            str_no_map[t_str].rec += (amount / number);
        }
    }
    for (int i = 0; i < N; ++i) {
        cout << arr_str[i] << " " << str_no_map[arr_str[i]].rec - str_no_map[arr_str[i]].giv << endl;
    }
}

int main()
{
    freopen("gift1.in", "r", stdin);
    freopen("gift1.out", "w", stdout);
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        cin >> arr_str[i];
        str_no_map[arr_str[i]] = myrecord[i];
    }
    Solve();
	return 0;
}

Friday the Thirteenth：數學知識就可解決。。。

/*
ID: Jming
PROG: friday
LANG: C++
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
using namespace std;
int N, cnt[13] = { 0, 31, 0, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int ans[7];

void Solve() {
    int end_year = 1900+N;
    int days, week = 1;
    for (int i = 1900; i < end_year; ++i) {
        for (int j = 1; j <= 12; ++j) {
            if (2 == j) {
                if (!(i%100)) {
                    if (!(i%400)) { days = 29; }
                    else { days = 28; }
                }else {
                    if (!(i%4)) { days = 29; }
                    else { days = 28; }
                }
            }else { days = cnt[j]; }
            ans[(week+5)%7]++;
            week = (week+days%7)%7;
        }
    }
    printf("%d ", ans[6]);
    for (int i = 0; i < 6; ++i) {
        printf("%d", ans[i]);
        if (i != 5) {
            printf(" ");
        }
    }
    printf("\n");
}

int main()
{
    freopen("friday.in", "r", stdin);
    freopen("friday.out", "w", stdout);
    scanf("%d", &N);
    Solve();
    return 0;
}

Broken Necklace：

由於(3<=N<=350)，故以O(n^2)複雜度解決也可AC。。。不過，AC後看到給的分析裏有一種O(n)的方法也學了一下。

O(n^2):

/*
ID: Jming
PROG: beads
LANG: C++
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;
int N;
string str;

void Solve() {
    int sum1, sum2;
    int ans = -1;
    for (int i = 0; i < N; ++i) {
        char ch;
        sum1 = 1;
        if ('w' == str[i]) {
            for (int j = i+1; j != i; j = (j+1)%N) {
                if ('w' != str[j]) {
                    ch = str[j];
                    break;
                }
            }
        }else { ch = str[i]; }

        int r = (i + 1)%N;
        while (r != i) {
            if (('w' == str[r]) || (str[i] == str[r])) {
                ++sum1;
                r = (r+1)%N;
            }
            else break;
        }
        r = (r-1 + N)%N;

        int l = (i-1+N)%N;

        int sum2 = 0;
        if (l != r) {
            if ('w' == str[l]) {
                for (int j = l; j != r; j = (j-1+N)%N) {
                    if ('w' != str[j]) {
                        ch = str[j];
                        break;
                    }
                }
            }else { ch = str[l]; }
        }
        while (l != r) {
            if (('w' == str[l]) || (ch == str[l])) {
                ++sum2;
                l = (l-1+N)%N;
            }
            else break;
        }
        ans = max(ans, sum1+sum2);
    }
    printf("%d\n", ans);
}

int main()
{
    freopen("beads.in", "r", stdin);
    freopen("beads.out", "w", stdout);
    scanf("%d", &N);
    cin >> str;
    Solve();
    return 0;
}

O(n)：

關鍵：

（1）將環轉換爲所給串重複的兩個（即str+str）

（2）作預處理，myleft[pos][ ]，表示在字符串pos位置之前（不包括pos位置的字符），連續的相同顏色的珠子數

                            myright[pos][ ]，表示從字符串pos位置向右（包括pos位置的字符），連續的相同顏色的珠子數

/*
ID: Jming
PROG: beads
LANG: C++
*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
using namespace std;
#define MAX(a, b) ((a)>(b)?(a):(b))
#define MIN(a, b) ((a)<(b)?(a):(b))

int N;
string str;
const int MAX_N = 705;
int myleft[MAX_N][2], myright[MAX_N][2];

void Solve() {
    // 0: b 1:r
    myleft[0][0] = myleft[0][1] = 0;
    for (int i = 1; i <= N; ++i) {
        if ('b' == str[i-1]) {
            myleft[i][0] = myleft[i-1][0] + 1;
            myleft[i][1] = 0;
        }
        else if ('r' == str[i-1]) {
            myleft[i][1] = myleft[i-1][1] + 1;
            myleft[i][0] = 0;
        }
        else {
            myleft[i][0] = myleft[i-1][0] + 1;
            myleft[i][1] = myleft[i-1][1] + 1;
        }
    }

    myright[N][0] = myright[N][1] = 0;
    for (int i = N-1; i >= 0; --i) {
        if ('b' == str[i]) {
            myright[i][0] = myright[i+1][0] + 1;
            myright[i][1] = 0;
        }else if ('r' == str[i]) {
            myright[i][1] = myright[i+1][1] + 1;
            myright[i][0] = 0;
        }else {
            myright[i][0] = myright[i+1][0] + 1;
            myright[i][1] = myright[i+1][1] + 1;
        }
    }
    int ans = 0;
    for (int i = 0; i < N; ++i) {
        ans = MAX(ans, MAX(myleft[i][0], myleft[i][1])+ MAX(myright[i][0], myright[i][1]));
    }
    ans = MIN(ans, N>>1);
    printf("%d\n", ans);
}

int main()
{
    freopen("beads.in", "r", stdin);
    freopen("beads.out", "w", stdout);
    scanf("%d", &N);
    cin >> str;
    str = str+str;
    N = N<<1;
    Solve();
    return 0;
}