UVa 10560 Minimum Weight

题意：砝码称重1~N，问最少要几个砝码。

思路：很容易得到b[i] = sum[i-1] * 2 + 1。因为sum[i-1]的都是已表示完备的，因此下一个需要多出sum[i-1] + 1。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;typedef unsigned long long ull;#define mem(a, n) memset(a, n, sizeof(a))#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 41 + 5, MOD = 1e9 + 7;LL M = 4e18;int T, cas = 0;LL n, m;int k;ull a[105], b[N], sum[N];void init() {    b[0] = 0; b[1] = 1, sum[1] = 1;    for(int i = 2; i < N; i ++) {        b[i] = 2 * sum[i-1] + 1;        sum[i] = sum[i-1] + b[i];    }}int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);// freopen("/Users/acpple/out.txt", "w", stdout);#endif        init();    while(scanf("%lld%d", &n, &k), n + k) {        ull num = 0;        for(int i = 0; i < k; i ++) { scanf("%llu", &a[i]); num = max(num, a[i]); }        int tot = lower_bound(sum, sum + N, n) - sum;        printf("%d", tot);        for(int i = 1; i <= tot; i ++) printf(" %llu", b[i]); puts("");        for(int i = 0; i < k; i ++) {            LL cnt = a[i];            int flag = 1;            for(int j = tot; j >= 0; j --) {                if(cnt == 0) break;                if(sum[j] < abs(cnt)) {                    if(flag) {                        cnt = b[j+1] - cnt;                        printf("%llu", b[j+1]);                        flag = 0;                    } else {                        printf("%c%llu", cnt < 0 ? '+' : '-', b[j+1]);                        if(cnt > 0) cnt -= b[j+1];                        else cnt += b[j+1];                    }                }            }            puts("");        }    }        return 0;}