USACO 1.1

USACO 1.2
主要说一下格式。
头部必打 /*
ID： beihai2013
PROG： (题目名字，题目有描述)
LANG： C++
*/
然后用文件流打开输入输出。

1.1.1
a+b

/*ID: beihai2013PROG: testLANG: C++*/#include <iostream>#include <fstream>#include <string>using namespace std;int main() {    ofstream fout ("test.out");    ifstream fin ("test.in");    int a, b;    fin >> a >> b;    fout << a+b << endl;    return 0;}

1.1.2
简单计算
/*
ID: beihai2013
PROG: ride
LANG: C++
*/

include

include

include

include

include

include

include

using namespace std;
const int MAXN = 1000;
char s1[MAXN], s2[MAXN];
int main()
{
// scanf(“%s %s”, s1, s2);
freopen(“ride.in”, “r”, stdin);
freopen(“ride.out”, “w”, stdout);
cin >> s1 >> s2;
int ans1 = 1, ans2 = 1;
for(int i = 0 ; i < (int)strlen(s1) ; i++)
ans1 = (ans1 * (s1[i] - ‘A’ + 1)) % 47;
for(int i = 0 ; i < (int)strlen(s2) ; i++)
ans2 = (ans2 * (s2[i] - ‘A’ + 1)) % 47;
if(ans1 == ans2) cout << “GO” << endl;
else cout << “STAY” << endl;

//scanf("%s", s1);return 0;

}
1.1.5
简单模拟。所有人的钱总和为0，一开始所有人的钱数量也为0。每次一个人会分钱给另外几个人，不能均分的钱就自己留着。

/*    ID: beihai2013    PROG: gift1    LANG: C++*/#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <iostream>#include <string>#include <algorithm>#include <map>using namespace std;const int MAXN = 20;char name[MAXN];char p[MAXN][MAXN];int val[MAXN];map<string,int>mm;int main(){    freopen("gift1.in", "r", stdin);    freopen("gift1.out", "w", stdout);    int n;    scanf("%d", &n);    for(int i = 1 ; i <= n ; i++){        scanf("%s", p[i]);        mm[p[i]] = i;        val[i] = 0;    }    while(scanf("%s", name) != EOF){        int sum, m;        int u = mm[name];        scanf("%d%d", &sum, &m);        if(m == 0){            continue;        }        for(int i = 0 ; i < m ; i++){            scanf("%s", name);            int v = mm[name];            val[v] += sum / m;        }        val[u] -= sum - sum % m;    }    for(int i = 1 ; i <= n ; i++){        printf("%s %d\n", p[i], val[i]);    }    return 0;}

1.1.6
愈来愈难了……
模拟时间，输出从1900年起，第N年的每个月第13日分别为星期6712345的数量
题目容易看错，注意给出的条件是1900.1.1为星期一，所以还要转化一下。
然后要把12月的天数放前面，因为遍历一年的时候是遍历这一年的1-12月，1月是从上一年的12月变化来的。

/*    ID: beihai2013    TASK: friday    LANG: C++*/#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <string>#include <algorithm>#include <iostream>using namespace std;const int MAXN = 400 + 5;int ans[MAXN][10];int mon[20] = {0, 31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30};int day(int year, int month){    if(month == 3){        if(year % 400 == 0) return 29;        else if(year % 100 != 0 && year % 4 == 0)   return 29;    }    return mon[month];}void init(){    memset(ans, 0, sizeof(ans));    int now = 5;    for(int i = 1 ; i < MAXN ; i++){        int year = 1900 + i - 1;        for(int j = 1 ; j <= 7 ; j++)   ans[i][j] = ans[i - 1][j];        for(int j = 1 ; j <= 12 ; j++){            now = (now + day(year, j)) % 7;            if(now == 0)    now = 7;            ans[i][now]++;//            printf("year = %d, month = %d, now = %d, day(year,j) = %d\n", year, j, now, day(year, j));//            system("pause");        }    }}int main(){    freopen("friday.in", "r", stdin);    freopen("friday.out", "w", stdout);    init();    int n;    scanf("%d", &n);    for(int i = 1 ; i <= 7 ; i++){        printf("%d", ans[n][i]);        if(i == 7)  printf("\n");        else    printf(" ");    }    return 0;}

1.1.7
神坑模拟。刚开始想用O（N）做然后就各种做不出来。
用O（N^2）做的，对每个点找能达到的最左端和最右端。但是还要分w和非w的情况讨论。

/*    ID: beihai2013    TASK: beads    LANG: C++*/#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <iostream>#include <string>#include <algorithm>using namespace std;const int MAXN = 1e3;char str[MAXN];int l[MAXN], r[MAXN];int main(){    freopen("beads.in", "r", stdin);    freopen("beads.out", "w", stdout);    int n;    scanf("%d", &n);    scanf("%s", str);    for(int i = n ; i < 2 * n ; i++)    str[i] = str[i - n];    str[n * 2] = '\0';    for(int i = 0 ; i < 2 * n ; i++){        if(str[i] != 'w'){            int now = i;            while(now >= 0 && (str[now] == 'w' || str[now] == str[i]))  now--;            l[i] = i - now;            now = i;            while(now < 2 * n && (str[now] == 'w' || str[now] == str[i]))   now++;            r[i] = now - i;        }        else{            int now = i;            while(now >= 0 && str[now] == 'w')  now--;            if(now >= 0){                char temp = str[now];                while(now >= 0 && (str[now] == 'w' || str[now] == temp))    now--;                l[i] = i - now;            }            else    l[i] = i - now;            now = i;            while(now < 2 * n && str[now] == 'w')  now++;            if(now < 2 * n){                char temp = str[now];                while(now < 2 * n && (str[now] == 'w' || str[now] == temp))    now++;                r[i] = now - i;            }            else    r[i] = -i + now;        }    }    int ans = 0;    for(int i = 0 ; i < 2 * n ; i++)    ans = max(ans, l[i] + r[i + 1]);    ans = min(ans, n);    printf("%d\n", ans);    return 0;}