HDU-4403 A very hard Aoshu problem 枚举

直接枚举等号所在位置，然后左右两边用两个MAP来记录组成和的个数，然后运用乘法原理得到答案。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
#include<string>
#define LL long long
using namespace std;char s[20];int len;map<long long,int>fmp, tmp;long long get(int x, int y) {long long ret = 0;for (int i = x; i <= y; ++i) {ret = ret * 10 + s[i] - '0';}    return ret;
}void cal(int a, int b, map<long long,int>&mp) {if (a == b) {++mp[s[a]-'0'];return;}int mask = 1 << (b-a), last;for (int i = 0; i < mask; ++i) {long long temp = 0;last = a;for (int j = 0; j < (b-a); ++j) {if (i & (1 << j)) {temp += get(last, a + j);last = a + j + 1;}}temp += get(last, b);++mp[temp];}
}long long solve(int x) {long long ret = 0;fmp.clear(), tmp.clear();map<long long, int>::iterator it;cal(0, x, fmp);cal(x+1, len-1, tmp);for (it = fmp.begin(); it != fmp.end(); ++it) {ret += (long long)(it->second) * (long long)tmp[it->first];}return ret;
}int main(  )
{long long ret;while (scanf("%s", s), s[0] != 'E') {ret = 0;len = strlen(s);for (int i = 0; i <= len-2; ++i) {// 枚举等号所在的位置ret += solve(i);}printf("%I64d\n", ret);}return 0;
}

转载于:https://www.cnblogs.com/Lyush/archive/2012/09/22/2698435.html

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