我竟然做了一题水题,浪费我感情啊,5分钟内2个AC(还是1A);1.0-1背包View Code #include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int w[22];bool dp[20000000];int main(){ int n, m, i, j, b; while(~scanf("%d%d",&n,&b)) { m=0; for(i=1;i<=n;i++) scanf("%d",&    Read More
posted @ 2012-08-10 20:05
To be an ACMan
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二分法:适用于单调函数,单调增或单调减三分法:适用于单峰凸性函数,如二次函数缩写 l:left ,r: right , m1:mid 1,m2:mid2三分法模板:#define eps 10e-6double cal(){}//计算题目所需要的值while(l+eps<r){ m1=l+(r-l)/3; m2=r-(r-l)/3; v1=cal(m1); v2=cal(m2); if(v1<v2)l=m1; else r=m2;}AC代码:View Code #include<stdio.h>#include<string.h>#include<ma    Read More
posted @ 2012-08-10 10:58
To be an ACMan
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