Codeforces Round #300 A B C

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A - Cutting Banner 给出一个字符串S，从中间删掉一部分子串T之后，剩下的S-T的串里面有没有“CODEFORCES”这个串。

string a = "CODEFORCES", b;int main(){    while( cin >> b )    {        bool f = 0;        int len = b.length();        for( int i = 0; i < len; ++i )        {            for( int j = i+1; j <= len; ++j )            {                if( b.substr(0, i) + b.substr( j ) == a )                {                    f = 1;                    break;                }            }        }        puts( f ? "YES" : "NO" );    }    return 0;}

B - Quasi Binary给出一个数字n，问能不能分解成只包含数字 1，0的数。

将所有的只包含0，1的数字预处理出来，然后将其看成是价值，花费看成1，那么就是一个完全背包了。

#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#define ls rt << 1#define rs rt << 1 | 1#define pi acos(-1.0)#define eps 1e-8typedef long long ll;const int inf = 0x3f3f3f3f;const int N = 1000010;vector <int> _01, res;int dp[N], path[N];int n;void dfs( int cur ){    if( cur > N )        return;    _01.push_back( cur );    dfs( cur * 10 + 1 );    dfs( cur * 10 );}void find_path( int x ){    if( !x )        return;    res.push_back( path[x] );    find_path( x - path[x] );}int main(){    _01.clear();    dfs( 1 );    sort( _01.begin(), _01.end() );    while( ~scanf("%d", &n) )    {        res.clear();        memset( dp, inf, sizeof( dp ) );        path[0] = dp[0] = 0;        int sz = _01.size();        for( int i = 0; i < sz; ++i )        {            if( _01[i] <= n )            {                for( int j = _01[i]; j <= n; ++j )                {                    if( dp[j] > dp[j - _01[i]] + 1 )                    {                        dp[j] = dp[j - _01[i]] + 1;                        path[j] = _01[i];                    }                }            }        }        printf("%d\n", dp[n]);        find_path( n );        for( int i = 0; i < res.size(); ++i )        {            printf("%d ", res[i]);        }        puts("");    }    return 0;}

C - Tourist's Notes给出n，m表示某些人一共爬了n天的山，其中有m份记录。每份记录表示di天在高度hi的地方。问合理情况（hi - hi-1 <= 1）下爬到最高的地方是多少米，不合理输出“IMPOSSIBLE”。

注意特殊处理d1和dn的情况，因为d1,dn不一定是第一天或最后一天

#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#define ls rt << 1#define rs rt << 1 | 1#define pi acos(-1.0)#define eps 1e-8typedef long long ll;const int inf = 0x3f3f3f3f;int n, m;int d, h;int main(){    while( ~scanf("%d%d", &n, &m) )    {        int cur, maxx, lastd;        bool OK = 0;        scanf("%d%d", &lastd, &cur);        maxx = lastd - 1 + cur;        m--;        while ( m-- )        {            scanf("%d%d", &d, &h);            if( OK )                continue;            if( abs( d-lastd) < abs(h-cur) )            {                ///printf("d: %d h: %d\n", d, h);                OK = 1;                continue;            }            int t = ( d - lastd );            maxx = max( maxx, ( t + cur - h ) / 2 + h );            lastd = d;            cur = h;        }        maxx = max( maxx, n - lastd + cur );        if( OK )            puts("IMPOSSIBLE");        else            printf("%d\n", maxx);    }    return 0;}

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