usaco3.2.2 Stringsobits

一 原题

Stringsobits
Kim Schrijvers

Consider an ordered set S of strings of N (1 <= N <= 31) bits. Bits, of course, are either 0 or 1.

This set of strings is interesting because it is ordered and contains all possible strings of length N that have L (1 <= L <= N) or fewer bits that are `1'.

Your task is to read a number I (1 <= I <= sizeof(S)) from the input and print the Ith element of the ordered set for N bits with no more than L bits that are `1'.

PROGRAM NAME: kimbits

INPUT FORMAT

A single line with three space separated integers: N, L, and I.

SAMPLE INPUT (file kimbits.in)

5 3 19

OUTPUT FORMAT

A single line containing the integer that represents the Ith element from the order set, as described.

SAMPLE OUTPUT (file kimbits.out)

10011

二 分析

一开始怀着侥幸心理想通过判断每个数里有多少个1，从1开始枚举到答案。但是数据有一个点是找第（INT_MAX+1）个数。。（也就是说变量I要定义成long long，因为这个又WA了好几次啊。。）正确做法是先动态规划求出长度为N的01串里含有至少L个1的串个数。然后依次判断目标串里每一位是填1还是0。

三 代码

运行结果：

USER: Qi Shen [maxkibb3]TASK: kimbitsLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.000 secs, 4188 KB]   Test 2: TEST OK [0.000 secs, 4188 KB]   Test 3: TEST OK [0.000 secs, 4188 KB]   Test 4: TEST OK [0.000 secs, 4188 KB]   Test 5: TEST OK [0.000 secs, 4188 KB]   Test 6: TEST OK [0.000 secs, 4188 KB]   Test 7: TEST OK [0.000 secs, 4188 KB]   Test 8: TEST OK [0.000 secs, 4188 KB]   Test 9: TEST OK [0.000 secs, 4188 KB]   Test 10: TEST OK [0.000 secs, 4188 KB]   Test 11: TEST OK [0.000 secs, 4188 KB]   Test 12: TEST OK [0.000 secs, 4188 KB]   Test 13: TEST OK [0.000 secs, 4188 KB]All tests OK.
Your program ('kimbits') produced all correct answers!  This is yoursubmission #6 for this problem.  Congratulations!

AC代码：

/*ID:maxkibb3LANG:C++PROG:kimbits*/#include<cstdio>const int MAX_N = 31;int N, L;long long I;long long dp[MAX_N + 1][MAX_N + 1];int main() {    freopen("kimbits.in", "r", stdin);    freopen("kimbits.out", "w", stdout);    scanf("%d%d%lld", &N, &L, &I);        dp[0][0] = 1;    for(int i = 1; i <= N; i++) {        for(int j = 0; j <= L; j++) {            if(j == 0)                dp[i][j] = dp[i - 1][j];            else if(j > i)                dp[i][j] = dp[i][i];            else if(j == i)                dp[i][j] = dp[i][j - 1] + 1;            else                dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];        }    }         for(int i = 1; i < N; i++) {        if(I <= dp[N - i][L]) {            printf("0");        }        else {            printf("1");            I -= dp[N - i][L];            L--;        }    }    if(I == 1) printf("0");    else printf("1");    printf("\n");    return 0;}