USACO Section 2.2

USACO Section 2.2
Preface Numbering

/*    ID: beihai2013    TASK: preface    LANG: C++*//*每次取最大的合法数的组合删去，类似于二进制表示*/#include <bits/stdc++.h>using namespace std;#define LL long longconst int MAXN = 3500 + 5;const int MAXM = 7 + 3;int v0[MAXM] = {1, 5, 10, 50, 100, 500, 1000};int v[MAXM * 2];int vis[MAXN];LL cnt[MAXN][MAXM];char str[] = "IVXLCDM";void init(){    for(int i = 1 ; i < MAXN ; i++) {        for(int j = 0 ; j < 7 ; j++) cnt[i][j] = 0;        int temp = i;        for(int j = 6 ; j >= 0 ; j--) {            while(temp >= v0[j]) {                temp -= v0[j];                cnt[i][j]++;            }            for(int k = j - 1 ; k >= max(j - 2, 0) ; k--) {                if(k % 2) continue;                int ok = 0;                while(temp >= v0[j] - v0[k]) {                    temp -= v0[j] - v0[k];                    cnt[i][j]++;                    cnt[i][k]++;                    ok = 1;                }                if(ok) break;            }        }    }//    for(int i = 1 ; i <= 5 ; i++) {//        printf("i = %d, ", i);//        for(int j = 0 ; j < 7 ; j++) printf("%I64d ", cnt[i][j]);//        puts("");//    }    for(int i = 1 ; i < MAXN ; i++) {        for(int j = 0 ; j < 7 ; j++) cnt[i][j] += cnt[i - 1][j];    }}int main(){    freopen("preface.in", "r", stdin);    freopen("preface.out", "w", stdout);    init();    int n;    while(scanf("%d", &n) != EOF) {        for(int i = 0 ; i < MAXM ; i++) {            if(cnt[n][i]) {                printf("%c %lld\n", str[i], cnt[n][i]);            }        }    }    return 0;}

Subset Sums

/*    ID: beihai2013    TASK: subset    LANG: C++*//*    首先判断有没有解，即所有数和的一半是否为整数    其次记忆化搜索*/#include <bits/stdc++.h>using namespace std;#define LL long longconst int MAXN = 40;int vis[MAXN];LL dp[(MAXN + 1) * MAXN / 2 + 1][MAXN];LL dfs(int rest, int mark){    if(rest == 0) return 1;    else if(rest < 0) return 0;    else if(dp[rest][mark]) return dp[rest][mark];    else if((mark + 1) * mark / 2 < rest) return 0;    else {        dp[rest][mark] = 0;        for(int i = mark ; i >= 1 ; i--) dp[rest][mark] += dfs(rest - i, i - 1);//        if(dp[rest][mark] != 0) printf("dp[%d][%d] = %lld\n", rest, mark, dp[rest][mark]);        return dp[rest][mark];    }}int main(){    freopen("subset.in", "r", stdin);    freopen("subset.out", "w", stdout);    int n;    while(scanf("%d", &n) != EOF) {        int temp = (n + 1) * n / 2;        if(temp % 2 == 0) printf("%lld\n", dfs(temp / 2, n) / 2);        else printf("0\n");    }    return 0;}

Runaround Numbers

/*    ID: beihai2013    TASK: runround    LANG: C++*//*    竟然是暴力……刚开始以为数据很少能不能预处理出所有合法数然后二分    跑了小数据发现合法数特别多，所以试了一发暴力    偷看了一组数据枪毙两分钟*/#include <bits/stdc++.h>using namespace std;bool check(unsigned int u){//    int debug = u == 147 ? 1 : 0;    if(u < 10) return true;    int a[10], cnt = 0;    while(u) {        a[cnt++] = u % 10;        if(u % 10 == 0) return false;        u /= 10;    }    for(int i = 0 ; i < cnt / 2 ; i++)        swap(a[i], a[cnt - 1 - i]);//    if(debug)//    for(int i = 0 ; i < cnt ; i++) printf("%d ", a[i]);    int vis[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0};    int now = 0, num = 1;    vis[a[now]] = 1;//    printf("cnt = %d\n", cnt);    while(num < cnt) {        now = (now + a[now]) % cnt;//        printf("now = %d, cnt = %d\n", now, num);        if(vis[a[now]] == 1) {//            if(debug) printf("num = %d, now = %d, a = %d\n", num, now, a[now]);            break;        }        else vis[a[now]] = 1;        num++;    }    if(num < cnt || (now + a[now]) % cnt != 0) return false;    else return true;}int main(){    freopen("runround.in", "r", stdin);    freopen("runround.out", "w", stdout);    unsigned int n = 1;    while(cin >> n) {        for(unsigned int u = n + 1; u <= 999999999; u++) {            if(check(u)) {                cout << u << endl;                break;            }        }        n++;    }    return 0;}

Party Lamps

/*    ID: beihai2013    TASK: lamps    LANG: C++*//*    一个操作使用两次相当于没有使用，总共四种操作    因此，用二进制枚举使用了哪些操作，得到对应的01串为可能合法的答案集合    此时，就要判断已经用了当前次数后，是否能“毫无改变”的达到最终状态    通常会想到周期为2的判断，但是也可能几个二进制枚举的组合使得一个01串不发生改变，预处理出来    此处写残，就是没把一个操作都没取的组合特判掉    最后答案用string输出方便排序，此处有坑就是会出现重复的string需要稍加判断*/#include <bits/stdc++.h>using namespace std;#define mp make_pair#define fi first#define se second#define pb push_backtypedef pair<int,int> pii;pii cal(int u){    if(u == 0) return mp(0, 1);    else if(u == 1) return mp(0, 2);    else if(u == 2) return mp(1, 2);    else return mp(0, 3);}const int MAXN = 100 + 5;const int MAXM = 4 + 2;bool stor[1 << MAXM][MAXN];int v[MAXN];int mul[MAXN], mulcnt;string out[1 << MAXM];int outcnt;bool check2(int c, int cnt){    int temp = c - cnt;//    printf("temp = %d\n", temp);    for(int i = 0 ; i < mulcnt ; i++) if(temp % mul[i] == 0) return true;    return false;}int main(){    freopen("lamps.in", "r", stdin);    freopen("lamps.out", "w", stdout);    int n, c;    while(scanf("%d%d", &n, &c) != EOF) {        for(int i = 0 ; i < n ; i++) v[i] = 0;        int u;        while(scanf("%d", &u) != EOF && u != -1) v[u - 1] = 1;        while(scanf("%d", &u) != EOF && u != -1) v[u - 1] = -1;        mulcnt = 0;        mul[mulcnt++] = 2;        outcnt = 0;        for(int i = 0 ; i < (1 << 4) ; i++) {//            printf("i = %d\n", i);            for(int j = 0 ; j < n ; j++) stor[i][j] = 1;            for(int j = 0 ; j < 4 ; j++) {                if((1 << j) & i) {                    pii p = cal(j);                    while(p.fi < n) {                        stor[i][p.fi] = (stor[i][p.fi] == 0 ? 1 : 0);                        p.fi += p.se;                    }                }            }            int cnt = 0;            int ok = 1;            for(int j = 0 ; j < n ; j++) if(stor[i][j] == 0) {ok = 0; break;}            if(ok && i) {                for(int j = 0 ; j < 4 ; j++) if((1 << j) & i) cnt++;                mul[mulcnt++] = cnt;            }        }//        puts("first");        int flag = 1;        for(int i = 0 ; i < (1 << 4) ; i++) {//            printf("i = %d\n", i);            int ok = 1;            for(int j = 0 ; j < n ; j++) {                if(v[j] == 1 && stor[i][j] == 0) ok = 0;                else if(v[j] == -1 && stor[i][j] == 1) ok = 0;                if(ok == 0) break;            }            if(ok == 0) continue;            int cnt = 0;            for(int j = 0 ; j < 4 ; j++) if((1 << j) & i) cnt++;//            puts("second");            if(cnt > c) continue;            else if(!check2(c, cnt)) continue;//            puts("third");            out[outcnt] = "";            for(int j = 0 ; j < n ; j++) out[outcnt] += stor[i][j] + '0';            outcnt++;            flag = 0;        }        sort(out, out + outcnt);        for(int i = 0 ; i < outcnt ; i++) {            if(i == 0 || out[i] != out[i - 1]) cout << out[i] << endl;        }        if(flag) puts("IMPOSSIBLE");    }    return 0;}/*24412 11 5 15 -1-1*/