A. Detectors
水题，判断有多少个点被若干个圆的并所覆盖，枚举下就行。
B. Simple Editor
简单的记事本模拟，用链表来实现，有四个操作，插入字符，删除，左移，右移。注意链表记录的时候，要记录方向边，也就是双向的链表。
删除这个操作的时候，忘了更新。
代码 #include < iostream > #include < string > #include < stdio.h > #include < string .h > #include < set > #include < algorithm > #include < vector > using namespace std; const int MAX = 1000005 ; int ind[MAX], top, p; int val[MAX], rev[MAX]; void insert( int v){ ind[top] = ind[p]; if (ind[p] != - 1 ) rev[ind[p]] = top; ind[p] = top; rev[top] = p; val[top ++ ] = v; if (ind[p] != - 1 ) p = ind[p];} void del(){ if (ind[p] != - 1 ) { ind[p] = ind[ind[p]]; rev[ind[p]] = p; }} int main(){ int T, c = 0 ; scanf( " %d " , & T); while (T -- ) { int n; top = 1 ; // ind[0]表示根节点 p = 0 ; memset(ind, - 1 , sizeof (ind)); memset(rev, - 1 , sizeof (rev)); scanf( " %d " , & n); for ( int i = 0 ; i < n; i ++ ) { char com[ 10 ]; scanf( " %s " , com); if (com[ 0 ] == ' I ' ) { // getchar(); scanf( " %s " , com); // insert(getchar()); insert(com[ 0 ]); } else if (com[ 0 ] == ' D ' ) { del(); } else if (com[ 0 ] == ' L ' ) { if (rev[p] != - 1 ) p = rev[p]; } else if (com[ 0 ] == ' R ' ) { if (ind[p] != - 1 ) p = ind[p]; } } printf( " Case #%d: " , ++ c); int x = ind[ 0 ]; while (x != - 1 ) { printf( " %c " , val[x]); x = ind[x]; } printf( " \n " ); }}C. Soul Combination
水题，把两个串按L长度剖分后，交替合并。
D. Subsequence's Sum
给定一个数列，和一个阈值K，求数列中连续子序列满足和小于K的个数。
枚举前n项和，然后二分查找子序列的左端，统计个数就行了。注意二分查找的时候，前n项和中可能有相等的值。
代码 #include < iostream > #include < string > #include < stdio.h > #include < string .h > #include < set > #include < algorithm > #include < vector > using namespace std;typedef long long int64; const int MAX = 1000005 ;int64 a[MAX], b[MAX]; int N, A, B, C, M, K;int64 go(){ int64 res = 0 ; b[ 1 ] = A, b[ 2 ] = B; for ( int i = 3 ; i <= N; i ++ ) { b[i] = (int64(b[i - 1 ]) * b[i - 2 ] + C) % M; } a[ 1 ] = b[ 1 ]; for ( int i = 2 ; i <= N; i ++ ) { a[i] = a[i - 1 ] + b[i]; } a[ 0 ] = 0 ; // for(int i = 1; i <= N; i++) printf("@%d\n", a[i]); for ( int i = 1 ; i <= N; i ++ ) { int low = 0 , high = i; while (low <= high) { int mid = (low + high) / 2 ; if (a[i] - a[mid] < (int64)K) high = mid - 1 ; else low = mid + 1 ; } res += (int64)i - low; } return res;} int main(){ int T, c = 0 ; scanf( " %d " , & T); while (T -- ) { scanf( " %d%d%d%d%d%d " , & N, & A, & B, & C, & M, & K); printf( " Case #%d: %lld\n " , ++ c, go()); }}E. SZX sensei's lunatic base
十进制与题目中的进(d1,d2,,,dn)制转换。
不理解进制的本质，若进制是(d1,d2,,,dn)，从低位到高位，则该数按该进制展开后是
X = a0 + a1 * d1 + a2 * d1 * d2 + a3 * d1 * d2 * d3 + ... + an * d1 * d2 * ... * dn
转载于:https://www.cnblogs.com/litstrong/archive/2011/01/02/1924033.html
