HDOJ 1203 I NEED A OFFER!(简单背包)
2013-12-25 13:33
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求得不到 offer 的最小概率即可。
显然可以转化为简单背包。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 10100, maxv = 10100;
int n, m;
int c[maxn];
double p[maxn];
double dp[maxv];
int main()
{
while(~scanf("%d%d", &m, &n) && (n || m)) {
for(int i = 0; i < n; i++) {
scanf("%d%lf", &c[i], &p[i]);
p[i] = 1.0 - p[i];
}
for(int i = 0; i <= m; i++) dp[i] = 1;
for(int i = 0; i < n; i++) {
for(int v = m; v >= 0; v--) {
if(v-c[i] >= 0)
dp[v] = min(dp[v], dp[v-c[i]]*p[i]);
}
}
double ans = (1-dp[m]) * 100;
printf("%.1lf%%\n", ans);
}
return 0;
}
显然可以转化为简单背包。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 10100, maxv = 10100;
int n, m;
int c[maxn];
double p[maxn];
double dp[maxv];
int main()
{
while(~scanf("%d%d", &m, &n) && (n || m)) {
for(int i = 0; i < n; i++) {
scanf("%d%lf", &c[i], &p[i]);
p[i] = 1.0 - p[i];
}
for(int i = 0; i <= m; i++) dp[i] = 1;
for(int i = 0; i < n; i++) {
for(int v = m; v >= 0; v--) {
if(v-c[i] >= 0)
dp[v] = min(dp[v], dp[v-c[i]]*p[i]);
}
}
double ans = (1-dp[m]) * 100;
printf("%.1lf%%\n", ans);
}
return 0;
}
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