UVA 10032 Tug of War
来源:互联网 发布:unity3d destroy怎么用 编辑:程序博客网 时间:2024/04/28 11:16
/*注意移位的时候要转化成64位的,因为1默认为32位的*/#include<stdio.h>#include<string>#define N 105#define sz(v) ((int)(v).size())#define rep(i, a, b) for (int i = (a); i < (b); ++i)#define repf(i, a, b) for (int i = (a); i <= (b); ++i)#define repd(i, a, b) for (int i = (a); i >= (b); --i)#define clr(x) memset(x,0,sizeof(x))#define clrs( x , y ) memset(x,y,sizeof(x))#define out(x) printf(#x" %d\n", x)#define sqr(x) ((x) * (x))typedef __int64 ll;int n,sum;int a[N];ll dp[45005];int max(int a,int b) {return a>b?a:b;}void solve() {int i,j;int m=sum/2;clr(dp);dp[0]=1;//dp[0]为1,即0个人的时候为1for(i=1;i<=n;i++) {for(j=m;j>=a[i];j--) {dp[j]|=dp[j-a[i]]<<1;//因为直接用值表示人数的话可能会被覆盖,所以用二进制标记}}int post=n/2;for(j=m;j>=0;j--) {if (n%2==0 && dp[j]&((ll)1<<post) ) break;if (n%2!=0 && (dp[j]&((ll)1<<(post+1)) || dp[j]&((ll)1<<post) )) break;}printf("%d %d\n",j,sum-j);}int main() {#ifndef ONLINE_JUDGEfreopen("in.txt","r",stdin);#endifint i;while(scanf("%d",&n)!=EOF) {sum=0;for(i=1;i<=n;i++) {scanf("%d",&a[i]);sum+=a[i];}solve();}return 0;}
0 0
- UVA 10032 Tug of War
- UVa 10032 - Tug of War
- UVA 10032(Tug of War)
- UVa Problem 10032 Tug of War (拔河)
- UVA - 10032 Tug of War (二进制标记+01背包)
- Tug of War
- Tug of War
- poj2576 tug of war
- Tug of War
- ZCMU-Tug of War
- ZOJ 1880 Tug of War
- ZOJ - 1880 Tug of War
- zoj 1880 - Tug of War
- POJ 2576 Tug of War
- POJ 2576 Tug of War
- lightoj 1147 - Tug of War
- ZOJ1880 POJ2576 Tug of War,DP
- (挑战编程_8_5)Tug of War
- dlmalloc 2.8.6 源码详解(2)
- 1.4插入排序-表插入排序
- Failed to install HelloWorld.apk on device 'emulator-5556': EOF
- jumper longest path --pending--
- Codeforces 484B Maximum Value(排序+二分)
- UVA 10032 Tug of War
- Java小品——简单的JDBC应用实现。
- 第10周项目4 大赛奖品
- poj解题报告——2209
- 老实的工程师,是否比较容易吃亏(转载)
- NES文件利用MATLAB可视化
- 不敢娶的媳妇
- SVN cleanup失败解决办法
- Android发post请求, 服务器端如何获取参数