Codeforces Round #197 (Div. 2)

A.Helpful Maths

分析：将读入的字符转化为数字，直接排个序就可以了。

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <algorithm>

using namespace std;



const int N = 500;

int seq[N];



int main() {

    char c;

    int t = 0, idx = 0;

    while ((c = getchar()) != EOF) {

        if (c == '+' || c == '\n') {

            seq[idx++] = t;

            t = 0;

        }

        else t = t * 10 + c - '0';

    }

    sort(seq, seq+idx);

    printf("%d", seq[0]);

    for (int i = 1;  i < idx; ++i) printf("+%d", seq[i]);

    puts("");

    return 0;

} 

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B.Xenia and Ringroad

分析：模拟即可，大于当前位置直接走，否则绕一圈后走到目标位置。

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <algorithm>

using namespace std;



typedef long long LL;

int n, m;



int main() {

    while (scanf("%d %d", &n, &m) != EOF) {

        int last = 1, x;

        LL ret = 0;

        while (m--) {

            scanf("%d", &x);

            if (x >= last) {

                ret += x - last;

                last = x;

            } else {

                ret += n-last+x;

                last = x;

            }

        }

        printf("%I64d\n", ret);

    }

    return 0;

}

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C.Xenia and Weights

分析：直接搜索即可，无法证明为何会如此快的找到答案或者退出。

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;



const int N = 1005;

char str[15];

int m;

int path[N];



bool dfs(int p, int left, int right, int last) {

    if (p == 0) return true;

    int delta = abs(left-right);

    for (int i = 1; i <= 10; ++i) {

        if (last == i) continue;

        if (str[i] && delta >= 0 && delta < i) {

            path[p] = i;

            if (left < right) {

                if (dfs(p-1, left+i, right, i)) return true;

            }

            else if (dfs(p-1, left, right+i, i)) return true;

        }

    }

    return false;

}



int main() {

    scanf("%s", str+1);

    scanf("%d", &m);

    for (int i = 1; i <= 10; ++i) str[i] -= '0';

    if (!dfs(m, 0, 0, 0)) {

        puts("NO");

    } else {

        puts("YES");

        for (int i = m; i > 0; --i) {

            printf(i == m ? "%d" : " %d", path[i]);

        }

        puts("");

    }

    return 0;

}

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D.Xenia and Bit Operations

分析：由于每次改变的位置只会改变与其相关位置的变化，对于其余部分的结果可以使用map存储起来，写的时候对log2(x)没有处理好，导致精度误差，教训啊。

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <map>

#include <cmath>

using namespace std;



const int N = (1<<17)+5;

int n, m;

int seq[N];

int cpy[N];

map<pair<int,int>, int>mp;

map<pair<int,int>, int>::iterator it;



int cal(int p, int l, int r) {

    if (r - l == 1) return mp[make_pair(l, r)] = (seq[l] | seq[r]);

    int mid = (r+l) >> 1;

    int f = (int)floor(log((r-l+1)*1.0)/log(2.0)+0.5); // 直接除取整会带来精度丢失 

    if (p <= mid) {

        if (f & 1) {// or

            return (mp[make_pair(l, mid)] = cal(p, l, mid)) | mp[make_pair(mid+1, r)];

        } else {

            return (mp[make_pair(l, mid)] = cal(p, l, mid)) ^ mp[make_pair(mid+1, r)];

        }

    } else {

        if (f & 1) {

            return mp[make_pair(l, mid)] | (mp[make_pair(mid+1, r)] = cal(p, mid+1, r));

        } else {

            return mp[make_pair(l, mid)] ^ (mp[make_pair(mid+1, r)] = cal(p, mid+1, r));

        }

    }

}



int main() {

    scanf("%d %d", &n, &m);

    int LIM = 1 << n;

    for (int i = 1; i <= LIM; ++i) {

        scanf("%d", &seq[i]);

        cpy[i] = seq[i];

    }

    for (int i = 1, k = 0; i < LIM; i <<= 1, ++k) {

        for (int j = 1; j+i <= LIM; j += (i<<1)) {

            if (k & 1) { // xor 

                cpy[j] ^= cpy[j+i];

                mp[make_pair(j, j+(i<<1)-1)] = cpy[j];

            } else { // or

                cpy[j] |= cpy[j+i];

                mp[make_pair(j, j+(i<<1)-1)] = cpy[j];

            }

        }

    }

    int p, b;

    while (m--) {

        scanf("%d %d", &p, &b);

        seq[p] = b;

        printf("%d\n", cal(p, 1, LIM));

    }

    return 0;

}

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