CF GYM 100703G Game of numbers

题意：给n个数，一开始基数为0，用这n个数依次对基数做加法或减法，使基数不超过k且不小于0，输出最远能运算到的数字个数，输出策略。

解法：dp。dp[i][j]表示做完第i个数字的运算后结果为j的可能性，可能为1，不可能为0，于是得到状态转移方程dp[i][j] = dp[i - 1][j - a[i]] | dp[i - 1][j + a[i]]，当被转移的状态不在0～k之间时特判。dp时记录路径。

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long
using namespace std;
bool dp[1005][1005];
bool path[1005][1005];
int a[1005];
int main()
{int n, k;while(~scanf("%d%d", &n, &k)){for(int i = 1; i <= n; i++)scanf("%d", &a[i]);memset(dp, 0, sizeof dp);memset(path, 0, sizeof path);dp[0][0] = 1;int ans = 0;int pos = 0;for(int i = 1; i <= n; i++){for(int j = 0; j <= k; j++){if(j - a[i] < 0 && j + a[i] > k)continue;if(j - a[i] < 0){dp[i][j] = dp[i - 1][j + a[i]];path[i][j] = 0;}else if(j + a[i] > k){dp[i][j] = dp[i - 1][j - a[i]];path[i][j] = 1;}else{dp[i][j] = (dp[i - 1][j - a[i]] | dp[i - 1][j + a[i]]);if(dp[i - 1][j - a[i]])path[i][j] = 1;}if(dp[i][j]){ans = i;pos = j;}}if(ans != i)break;}printf("%d\n", ans);vector <bool> v;for(int i = ans; i > 0; i--){v.push_back(path[i][pos]);if(path[i][pos])pos -= a[i];elsepos += a[i];}int len = v.size();for(int i = len - 1; i >= 0; i--){if(v[i])printf("+");elseprintf("-");}puts("");}return 0;
}

转载于:https://www.cnblogs.com/Apro/p/4685436.html

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