NWERC 2011 ABCDEH 题解

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A：SPOJ NWERC11A A - Binomial coefficients

题解：点击打开链接

B: 点击打开链接

Bird tree

从下到上发现是个gcd的过程（辗转相除

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int main() {    int T;    scanf("%d", &T);    while (T--) {        int a, b;        scanf("%d/%d", &a, &b);        while (a - b) {            if (a > b) {                putchar('R');                a -= b;            } else {                putchar('L');                b -= a;            }            swap(a, b);        }        puts("");    }    return 0;}

C:点击打开链接

Movie collection

#include <cstdio>#include <algorithm>typedef long long ll;const int N = 100005 * 3;int pos[N], a[N], nn;void clear() {for(int i = 0; i <= nn; i ++) {a[i] = 0;}}void update(int idx, int val) {    while( idx <= nn ) {    a[idx] += val;idx += idx & (-idx);    }}int sum(int idx) {int s = 0;while(idx > 0) {s += a[idx];idx -= idx & (-idx);}return s;}int main() {int T;scanf("%d", &T);int n, m, x;while(T-- > 0) {scanf("%d%d", &n, &m);nn = n + m;clear();for(int i = 1; i <= n; i ++) {pos[i] = i+m;update(i+m, 1);}for(int i = m; i > 0; i --) {scanf("%d", &x);printf("%d", sum(pos[x]) - 1);if(i > 1) printf(" ");update(pos[x], -1);update(i, 1);pos[x] = i;}puts("");}return 0;}

D:题目链接： http://www.spoj.com/problems/NWERC11D/

Piece it together

题解：点击打开链接

E: 点击打开链接

Please, go first

#include <cstdio>#include <algorithm>#include<iostream>#include<string.h>#include<vector>using namespace std;const int N = 300;vector<int>G[N];char s[30000];int main() {    int cas, idx, u, Q, n;    scanf("%d", &cas);    while (cas -- >0) {        scanf("%d",&n);        scanf("%s",s);        for(int i = 0; i < N; i++)G[i].clear();        for(int i = 0; i < n; i++) {            G[s[i]].push_back(i);        }        int ans = 0;        for(int i = 0; i < N; i++)if(G[i].size()){            for(int j = 0; j < G[i].size()-1; j++) {                ans += G[i][G[i].size()-1] - G[i][j];            }        }        for(int i = 0; i < N; i++)            if(G[i].size())        {            int len = G[i].size();            len--;            ans -= (1+len)*len/2;        }        cout<<ans*5<<endl;    }    return 0;}/*992AB2AA6ABABAB10AAAAAAAAAA1A5ABCDD5ABCDA*/

H:点击打开链接

Tichu

题解：http://blog.csdn.net/qq574857122/article/details/37778285

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