标签:txt isp inpu 百度 mil stdout bsp ons target
题意 : 给出若干个物品的数量和单个的重量、问你能不能刚好组成总重 S
分析 :
由于物品过多、想到二进制优化
其实这篇博客就是存个二进制优化的写法
关于二进制优化的详情、百度一下有更多资料
#include<bits/stdc++.h> #define LL long long #define ULL unsigned long long #define scl(i) scanf("%lld", &i) #define scll(i, j) scanf("%lld %lld", &i, &j) #define sclll(i, j, k) scanf("%lld %lld %lld", &i, &j, &k) #define scllll(i, j, k, l) scanf("%lld %lld %lld %lld", &i, &j, &k, &l) #define scs(i) scanf("%s", i) #define sci(i) scanf("%d", &i) #define scd(i) scanf("%lf", &i) #define scIl(i) scanf("%I64d", &i) #define scii(i, j) scanf("%d %d", &i, &j) #define scdd(i, j) scanf("%lf %lf", &i, &j) #define scIll(i, j) scanf("%I64d %I64d", &i, &j) #define sciii(i, j, k) scanf("%d %d %d", &i, &j, &k) #define scddd(i, j, k) scanf("%lf %lf %lf", &i, &j, &k) #define scIlll(i, j, k) scanf("%I64d %I64d %I64d", &i, &j, &k) #define sciiii(i, j, k, l) scanf("%d %d %d %d", &i, &j, &k, &l) #define scdddd(i, j, k, l) scanf("%lf %lf %lf %lf", &i, &j, &k, &l) #define scIllll(i, j, k, l) scanf("%I64d %I64d %I64d %I64d", &i, &j, &k, &l) #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 #define lowbit(i) (i & (-i)) #define mem(i, j) memset(i, j, sizeof(i)) #define fir first #define sec second #define VI vector<int> #define ins(i) insert(i) #define pb(i) push_back(i) #define pii pair<int, int> #define VL vector<long long> #define mk(i, j) make_pair(i, j) #define all(i) i.begin(), i.end() #define pll pair<long long, long long> #define _TIME 0 #define _INPUT 0 #define _OUTPUT 0 clock_t START, END; void __stTIME(); void __enTIME(); void __IOPUT(); using namespace std; const int maxn = 1e6 + 10; const LL mod = 1e9 + 7; int num = 0; LL w[maxn]; LL dp[maxn]; int main(void){__stTIME();__IOPUT(); int nCase; sci(nCase); while(nCase--){ num = 0; int n, q; scii(n, q); for(int i=0; i<n; i++){ int V, C; scii(V, C); int tot = (1<<C)-1; int j; for(j=1; j<=tot; j<<=1) w[num++] = 1LL * j * (LL)V % mod, tot -= j; if(tot > 0){ LL rem = tot; w[num++] = 1LL * rem * (LL)V % mod; } } mem(dp, 0); dp[0] = 1; for(int i=0; i<num; i++) for(int j=(int)1e4; j>=w[i]; j--) dp[j] = (dp[j] + dp[j-w[i]]) % mod; while(q--){ int num; sci(num); printf("%lld\n", dp[num]); } } __enTIME();return 0;} void __stTIME() { #if _TIME START = clock(); #endif } void __enTIME() { #if _TIME END = clock(); cerr<<"execute time = "<<(double)(END-START)/CLOCKS_PER_SEC<<endl; #endif } void __IOPUT() { #if _INPUT freopen("in.txt", "r", stdin); #endif #if _OUTPUT freopen("out.txt", "w", stdout); #endif }
2018 焦作网络赛 K Transport Ship ( 二进制优化 01 背包 )
标签:txt isp inpu 百度 mil stdout bsp ons target
原文地址:https://www.cnblogs.com/Rubbishes/p/9676688.html